An Introduction to Sieve Methods and their applications has the following Theorem 6.3.4
$$|\lbrace p\leq x:\;|p+\alpha| \text{ is prime} \rbrace|<\frac{cx}{(\log x)^2} \prod_{p\mid\alpha}\left(1-\frac{1}{p}\right)^{-1}$$
The proof does this by showing the quantity we wish to estimate is less than $ z +S(\mathcal{A},\mathcal{P},z)$ where $\mathcal{P}$ is the set of all primes and $\mathcal{A}=\lbrace n+\alpha:\; n\leq x\rbrace$
Then Brun's Sieve implies that $S(\mathcal{A},\mathcal{P},z)\ll xW(z) + O(z^\theta)$ for some constant $\theta$.
Now is where is have the problem, the last line is the problem, It is supposedly by Merten's theorem by I do not see the connection.
\begin{align*} W(z)&=\prod_{p<z}\left(1-\frac{\omega(p)}{p}\right)\\ &=\prod_{p<z, p \mid \alpha}\left(1-\frac{1}{p}\right)\prod_{p<z, p\nmid\alpha}\left(1-\frac{2}{p}\right)\\ &=\prod_{p<z, p|\alpha}\left(1-\frac{1}{p}\right)\left(1-\frac{2}{p}\right)^{-1}\prod_{p<z}\left(1-\frac{2}{p}\right)\\ &\leq\frac{C}{(\log z)^2}\prod_{p\mid \alpha}\left(1-\frac{1}{p}\right)^{-1}\\ \end{align*}
It then immediately states (I dont understand this either)
$$|\lbrace p\leq x:\;|p+\alpha| \text{ is prime} \rbrace|<\frac{x}{(\log z)^2} \prod_{p<z} \left(1-\frac{2}{p}\right)+z^\theta$$
and that the theorem follows by taking $z=x^{1/2 \theta}$ (Again I do not understand)
I think there is a msitake in the book, this is what I have arrived at
\begin{align*} W(z)&=\prod_{p<z}\left(1-\frac{\omega(p)}{p}\right)\\ &=\prod_{p<z, p|\alpha}\left(1-\frac{1}{p}\right)\prod_{p<z, p\nmid\alpha}\left(1-\frac{2}{p}\right)\\ &\leq \prod_{p<z, p|\alpha}\left(1-\frac{1}{p}\right)\prod_{p<z, p\nmid\alpha}\left(1-\frac{1}{p}\right)^2\\ &=\left(\prod_{p<z}\left(1-\frac{1}{p}\right)\right)^2\prod_{p<z, p|\alpha}\left(1-\frac{1}{p}\right)^{-1}\\ &\leq\frac{C}{(\log{z})^2}\prod_{p|\alpha}\left(1-\frac{1}{p}\right)^{-1}\\ \end{align*} Where the last line is by Merten's theorem
Therefore
$$ \lbrace p\leq x:\;|p+\alpha| \text{is prime} \rbrace|<\frac{Cx}{(\log z)^2}\prod_{p|\alpha}\left(1-\frac{1}{p}\right)^{-1}+O(z^\theta)+z $$
Take $z=x^{1/2\theta}$ and observe that $x^{1/2\theta}<\sqrt{x}$ and that there exists a positive constant $A$ such that $A\sqrt{x}<\frac{x}{(\log x)^2}$ for all $x>1$.