I want to ask the estimate of $$P_d(x)=\text{card}\{ p\in\mathbb{P} \mid x=p+dq \ \ \text{ for some } q\in \mathbb{P}\},$$ where $d$ is a given positive integer(or a sufficiently small positive integer).
I think the magnitude be given by $$P_d(x)=\frac{x}{\log^2 x}\bigg(1+O\Big(\frac{1}{\log x}\Big)\bigg)f(d)$$ for some $f(d)$ which has an arithmetic character like the product term of the Hardy-Littlewood asymptotic for the binary Goldbach conjecture, or the term in the Chen's theorem for the Goldbach conjecture. But the point of the quetion is that the magnitude be given by $x\log^{-2}x$, I ask this question because I didn't make it sure.
It might be considered via the estimate (especially its upper bound) of $\text{card}\{ p\in\mathbb{P} \mid \Omega(x-p)\leq k \}$, where the case of $k=2$ is studied but I had not found for $k>2$. And additionally, if we consider the problem with the sum $\sum_{n\leq x, \ \Omega(n)=k}1$ or $\sum_{n\leq x, \ \omega(n)=k}1$, then the term $(\log\log x)^{k-1}$ is contained in their estimates $(\log\log x)^{k-1}x\log^{-1} x$.
Several facts that I considered are as follows:
As I said above, Chen's theorem shows that the number of representations of $x$ as a sum of a prime and a product of at most two primes is $x\log^{-2}x$ in its magnitude. Hence if $d\in\mathbb{P}$ then, first, it is obvious that $P_d(x)\ll x\log^{-2}x\prod_{p|x}(p-1)/(p-2) $. But the problem is for an arbitrarily given $d$.
Another thinkable property is the Brun-Titchmarsh theorem which states that $$\pi(x,d,a)\leq \frac{2x}{\varphi(d)\log (x/d)},$$ where $\pi(x,d,a)$ count the primes congruent to $a$ modulo $d$ up to $x$, and where $\varphi(d)$ is the Euler's totient function. (Considering the range of $d$, I considered this statement rather than another well known theorems which improve the inequality.) Then now we consider the expression, with given $x$ and $d$, $$ x=p+dN, $$ where $N$ is an odd positive integer. By putting $N=1,3,5,\ldots, k \ (<x/d)$, the number of the expressions above is less than $2x/\varphi(d)\log (x/d)$, and then by considering the density of the number of the primes under $x/d$, I have thought that $$P_d(x)\leq \frac{2x}{\varphi(d)(\log x/d)^2}\bigg(1+O\Big(\frac{1}{\log x}\Big)\bigg)$$ which specifies $f(d)$. But of course it is not correct yet because it needs proof that the set of $N$ making $p\in \mathbb{P}$ also has the density $\log (x/d)$, it is the main problem in this discussion regardless of the choice of the theorem concerning $\pi(x,d,a)$.
Thanks for reading my question.
I have answered this question myself.
$\mathrm{Proposition.}$ Let $x\in 2\mathbb{N}$ and let $d$ be a given integer such that $(x,d)=1$. Then, \begin{align*} \mathrm{card}\{ p \in \mathbb{P} \mid x-dp \in \mathbb{P} \} \ll\frac{x/\varphi(d)}{(\log x)^2}\prod_{p|x}\bigg(1+\frac{1}{p}\bigg), \end{align*} where $\varphi(x)$ is the Euler's totient function.
$\mathrm{Proof.}$ The proof is essentially same to the proof of the statement $G(x)\ll {x}{\log^{-2} x}\prod_{p|x}(1+p^{-1})$ by using the Selberg's sieve, but there are several modifications for the multiplicative function $g$ and the set $\mathcal{P}$. Let $$A=\{a_n\mid a_n=n(x-dn), n\in \mathbb{N} \text{ with } n \leq x/d \}.$$ Let $\mathcal{F}^y$ be the set of prime factors of $y$ and let $\mathcal{P}^y=\mathbb{P}\setminus\mathcal{F}^y$. Let $2\leq z \leq \sqrt x$ and let $\mathcal{S}(A,\mathcal{P}^x,z)$ denote the number of elements of $A$ that are divisible by no prime $p\in \mathcal{P}^d$ such that $p<z$. If $\sqrt x < n < \lfloor x/d \rfloor-\sqrt x$ and if $a_n\equiv 0 \ (\text{mod } p)$ for some $p\in \mathcal{P}^d$ such that $p<z$, then $a_n=n(x-dn)$ is composite. Thus $$\mathrm{card}\{ p \in \mathbb{P} \mid x-dp \in \mathbb{P}\} \leq 2\sqrt x + \mathcal{S}(A,\mathcal{P},z)\leq 2\sqrt x + \mathcal{S}(A,\mathcal{P}^d,z),$$ our interest be an upper bound of $\mathcal{S}(A,\mathcal{P}^d,z)$.
$\ \ $Let $A_\delta$ be the set of elements of $A$ that are divisible by $\delta$. Then, $|A_\delta|=|A|g(\delta)+r(\delta)$, where $g$ is completely multiplicative function defined by \begin{align*} g(p)=& \begin{cases} 1/p \ \ \text{if } p\mid x ,\\ 2/p \ \ \text{if } p\nmid x , \end{cases} \end{align*} for $p\in\mathcal{P}^d$, and $r(\delta)$ is the remainder. Note that if we take $\mathbb{P}$ rather than $\mathcal{P}^d$ then $g$ is not a multiplicative function. Then by the Selberg's theorem, \begin{align} \mathcal{S}(A,\mathcal{P}^x,z) \ll \frac{x/d}{G(z)}+\sum_{\delta<z^2, \ \delta|P(z)}3^{\omega(\delta)}|r(\delta)|, \label{133} \end{align} where $P(z)=\prod_{p\in\mathcal{P}^d}p$ and $$G(z)= \sum_{1\leq m < z \atop m \in \langle\mathcal{P}^d\rangle}g(m),$$ where $\langle S \rangle=\{s_i^{a_i}\mid s_i\in S\cup\{1\}, \ a_i\in\mathbb{N}\}$.
$ \ \ $ We first estimate $G(z)$. Let $m=\prod_{i=1}^{k} f_i^{\alpha_i}\prod_{j=1}^{l} p_j^{\beta_j}$, where $f_i$ and $p_j$are classified by the conditions $f_i| x$ and $p_j\nmid x$, respectively, note that they are relatively prime to $d$. Then, first, $g(m)=m^{-1}\prod_{j=1}^{l} 2^{\beta_j}$. Now, let $d_0(m,y)$ denote the number of positive divisors of $m$ that are relatively prime to $y$, where $y$ is a positive integer. Since $d_0(m,xd)=\prod_{j=1}^{l} (\beta^j+1)\leq \prod_{j=1}^{l} 2^{\beta_j}$, we have $g(m)\geq {d_0(m,xd)}/{m} $. Then, \begin{align} G(z)\prod_{p|xd}\bigg(1-\frac{1}{p}\bigg)^{-1} &\geq \sum_{m<z\atop m \in \langle\mathcal{P}^d\rangle}\frac{d_0(m,xd)}{m}\bigg(\prod_{p|xd}\bigg(1-\frac{1}{p}\bigg)^{-1}\bigg) \nonumber \\ &=\sum_{m<z \atop m \in \langle\mathcal{P}^d\rangle}\frac{d_0(m,xd)}{m}\bigg(\sum_{t\in \langle\mathcal{F}^{xd}\rangle}\frac{1}{t}\bigg) \nonumber \\ &=\sum_{m<z \atop m \in \langle\mathcal{P}^d\rangle}d_0(m,xd)\bigg(\sum_{t\in \langle\mathcal{F}^{xd}\rangle}\frac{1}{mt}\bigg) \nonumber \\ &=\sum_{w=1}^{\infty}\frac{1}{w}\bigg(\sum_{m<z, \ m|w \atop m \in \langle\mathcal{P}^d\rangle, \ (w/m)\in \langle\mathcal{F}^{xd}\rangle }d_0(m,xd)\bigg) \geq\sum_{1\leq w<z}\frac{1}{w}\bigg(\sum_{ m|w \atop m \in \langle\mathcal{P}^d\rangle, \ (w/m)\in \langle\mathcal{F}^{xd}\rangle}d_0(m,xd)\bigg), \label{139} \end{align} note that we take the condition $p|xd$ for the product term rather than $p|x$ in order to let $w\in \mathbb{N}$ without any condition. Now we evaluate the bracket term above, let $w=\phi\prod_{i=1}^{k} f_i^{\gamma_i}\prod_{j=1}^{l} p_j^{\delta_j}$, where $\phi$ is the greatest integer satisfying $\phi|d$ and $f_i,p_j$ are classified by the conditions $p_i| x$ and $q_j\nmid x$, respectively. Since $w/m\in \langle\mathcal{F}^{xd}\rangle$ and $m \in \langle\mathcal{P}^d\rangle$, every $m$ has the form $m=\prod_{i=1}^{k} f_i^{\alpha_i}\prod_{j=1}^{l} p_j^{\delta_j}$. Then, we first have $d_0(m,xd)=\prod_{j=1}^{l}(\delta_j+1)$. And since the number of such divisors $m$ is equal to $\prod_{i=1}^{k}(\gamma_i +1)$, we see that \begin{align*} \sum_{ m|w \atop m \in \langle\mathcal{P}^d\rangle, \ (w/m)\in \langle\mathcal{F}^{xd}\rangle}d_0(m,xd) =\prod_{i=1}^{k}(\gamma_i +1)\prod_{j=1}^{l}(\delta_j +1)=d_0\Big(\frac{w}{\phi}\Big). \end{align*} Thus, the summation involving the sum above be given by $$\sum_{1\leq w<z}\frac{d_0(w,d)}{w},$$ and we see that \begin{align*} \sum_{1\leq w<z}\frac{d_0(w,d)}{w}>\prod_{p|d}\bigg(1+\sum_{k=1}\frac{k+1}{p^k}\bigg)^{-1}\sum_{1\leq w<z}\frac{d_0(w)}{w}=\prod_{p|d}\bigg(\frac{p}{p-1}\bigg)^{-2}\sum_{1\leq w<z}\frac{d_0(w)}{w} \end{align*} since $d_0(w)$ is a multiplicative function generated by $\prod_{p\in \mathbb{P}}(1+\sum_{k=1}(k+1)p^{-k})$. Therefore, \begin{align*} G(z) &\geq \prod_{p|xd}\bigg(1-\frac{1}{p}\bigg)\prod_{p|d}\bigg(\frac{p}{p-1}\bigg)^{-2}\sum_{1\leq w < z }\frac{d_0(w)}{w}\\ &=\prod_{p|x}\bigg(1-\frac{1}{p}\bigg)\prod_{p|d}\bigg(\frac{p}{p-1}\bigg)^{-1}\sum_{1\leq w < z }\frac{d_0(w)}{w}. \end{align*} Since $\sum_{1\leq w<z}{d_0(w)}{w}^{-1}=2^{-1}(\log z)^2+ O(\log z)$, we have an upper bound \begin{align} \frac{x/d}{G(z)} &\leq 2\frac{x/d}{(\log z)^2} \prod_{p|x}\bigg(1-\frac{1}{p}\bigg)^{-1} \prod_{p|d}\bigg(\frac{p}{p-1}\bigg) \Big(1+O(\log^{-1} z)\Big), \nonumber \\ &=2\frac{x/\varphi(d)}{(\log z)^2}\prod_{p|x}\bigg(1-\frac{1}{p}\bigg)^{-1}\Big(1+O(\log^{-1} z)\Big) \label{132} \end{align} where $z$ is undetermined yet.
$ \ \ $ For the remainder, since $|r(\delta)|\leq 2^l\leq 2^{\omega(\delta)}<\delta$, then \begin{align} \sum_{\delta<z^2 \atop \delta|P(z)}3^{\omega(\delta)}|r(\delta)| \leq \sum_{\delta<z^2 }6^{\omega(\delta)} =\sum_{\delta<z^2 }2^{(\log 6/\log 2)\omega(\delta)} \leq \sum_{\delta<z^2 }\delta^{(\log 6/\log 2)}<z^{2+2(\log 6/\log 2)}. \label{131} \end{align}
$ \ \ $Now we let $z=x^{1/N}$ on the two results above, where $N$ is suitable constant. Indeed we can evaluate that the remainder is less than $x^{9/10}$ if $N=8$, and is less than $x^{3/5}$ if $N=12$. Consequently, \begin{align*} \mathcal{S}(A,\mathcal{P}^d,z)\leq c\frac{x/\varphi(d)}{(\log x)^2}\prod_{p|x}\bigg(1-\frac{1}{p}\bigg)^{-1}\Big(1+O(\log^{-1} x)\Big)+x^{b} \end{align*} for some positive constants $c$ and $b(<1)$ depending on $N$. And by comparing the product parts, we prove the proposition.