Let $0<s<1<p<\infty$ and assume that $\phi\in L^p_{loc}(\mathbb{R}^N)$ such that $$ \|\phi\|_1:=\int_{\mathbb{R}^N}\frac{|\phi(y)|^p}{1+|y|^{N+sp}}\,dy<\infty. $$ Let $E\subset\mathbb{R}^N$ be a bounded subset and define for $\phi\in L^p_{loc}(\mathbb{R}^N)$ and $x\in E$ $$ \|\phi\|_2:=\int_{\mathbb{R}^N\setminus E}\frac{|\phi(y)|^{p}}{|x-y|^{N+sp}}\,dy. $$ Here $|E|$ denotes the Lebesgue meaure of $E$. Then we have $$ \|\phi\|_2\leq C\|\phi\|_1, $$ for some constant $C$ depending on $n,p,s$ and $|E|$. This type of question is posted on the link: Integral estimate in $\mathbb{R}^N$
But the above question asks something more.
Any help would be very much appreciated. Thanks.
Essentially what you want to prove is that if $x \in E$ then there exists C such that $$ \frac 1 {\vert x - y \vert^{N+sp}} \leqslant \frac C {1+\vert y \vert^{N+sp}} \tag{$\ast$}$$ for all $y \in \mathbb{R}^N \backslash E$. You say that you want C independent of $x$, but as it stands this is not possible. If $C$ is independent of $x$ then by taking $x$ close to $\partial E$ and $y$ close to $x$ the LHS of ($\ast$) becomes arbitrarily large while the RHS stays finite. To fix this issue you should take $x \in E' \subset \subset E$. Then $C$ will depend on $n$, $s$, $p$, $E$ and $E'$ (but not $x$).
You could probably prove ($\ast$) by some smart inequalities, but the simplest way (and also more useful for harder situations) is to check the asymptotics. Here a way to do this rigorously: Let $f :\overline{E'} \times \mathbb{R}^N \backslash E \to \mathbb{R}$ be given by $$ f(x,y) = \frac{1+\vert y \vert^{N+sp}}{\vert x -y \vert^{N+sp}}.$$ (Note $f$ is well-defined since $x \in E' \subset \subset E$) Let $M$ be the supremum of $f$ over $E' \times \mathbb{R}^N \backslash E$ and $\{(x_k,y_k)\}$ be a sequence such that $f(x_k,y_k) \to M$ as $k \to \infty$. Since $E'\subset \subset E$, after possibly passing to subsequence, $x_k \to \bar{x} \in \overline{E'}$.
Next, observe that after possibly passing to a subsequence, there are two options for $\{y_k\}$: either $\{y_k\}$ is unbounded or $\{y_k\}$ converges. Indeed, if $\{y_k\}$ is bounded then by compactness it has a convergent subsequence.
If $\{y_k\}$ converges to some $\bar{y} \in \mathbb{R}^n \backslash E$ then $M = f(\bar{x},\bar{y})<\infty$. If $\{y_k\}$ is unbounded then write $\hat{y}_k = y_k / \vert y_k \vert$ and $$f(x_k,y_k) = \frac{\vert y_k \vert^{-N-sp}+\vert \hat{y}_k \vert^{N+sp}}{\vert x/\vert y_k \vert -\hat{y}_k \vert^{N+sp}} =\frac{\vert y_k \vert^{-N-sp}+1}{\vert x/\vert y_k \vert -\hat{y}_k \vert^{N+sp}}. $$ Since $\hat{y}_k \in \partial B_1$ which is compact, again after passing to a subsequence $\hat{y}_k \to \omega \in \partial B_1$. Hence, taking $ k \to \infty$ gives $$f(x_k,y_k) \to \frac 1 {\vert \omega \vert^{N+sp}}=1. $$
Thus, we have that $f$ is bounded which implies ($\ast$). Finally, using ($\ast$) $$ \| \phi\|_2 = \bigg ( \int_{\mathbb{R}^N \backslash E} \frac{\vert \phi (y) \vert^p}{\vert x- y \vert^{N+sp}} d y \bigg )^\frac 1 p \leqslant C\bigg ( \int_{\mathbb{R}^N \backslash E} \frac{\vert \phi (y) \vert^p}{1+ \vert y \vert^{N+sp}} d y \bigg )^\frac 1 p \leqslant C\| \phi \|_1$$ (For $\| \cdot \|_1$ and $\| \cdot \|_2$ to be norms you need to take the $p$-th root of what you've written.)