Answer
Yes.
Work Step by Step
We will prove this is true. So Let $A$ and $B$ be two sets, and suppose $\mathcal{P}(A) = \mathcal{P}(B)$. To show $A=B$, we will show $A\subseteq B$ and $B\subseteq A$. So let $x\in A$, then $\{x\}\subseteq A$. This means $\{x\}\in \mathcal{P}(A)$. Now since $\mathcal{P}(A) = \mathcal{P}(B)$, we have $\{x\}\in \mathcal{P}(B)$, and, therefore, $\{x\}\subseteq B$. But this means every element in $\{x\}$ is in $B$. Thus $x\in B$, and ,therefore, $A\subseteq B$.
To show $B\subseteq A$, we use an analogous argument. So let $x\in B$. then $\{x\}\subseteq B$. This means $\{x\}\in \mathcal{P}(B)$. Now since $\mathcal{P}(B) = \mathcal{P}(A)$, we have $\{x\}\in \mathcal{P}(A)$, and, therefore, $\{x\}\subseteq A$. But this means every element in $\{x\}$ is in $A$. Thus $x\in A$, and ,therefore, $B\subseteq A$. Hence $A=B$.
