Let $x$ be an arbitrary element of $A$. then, there exists a subset say $X$, of set $A$ such that $x \in X$
Now, $X \subset A \Rightarrow X \in P(A)$
$\begin{array}{lc}\Rightarrow X \in P(B) & {[\because P(A)=P(B)]} \\ \Rightarrow X \subset B \Rightarrow x \in B & {[\because x \in X \text { and } X \subset B \Rightarrow x \in B]}\end{array}$
Thus, $x \in A \Rightarrow x \in B \quad \therefore A \subset B \quad \ldots (i)$
Now, let $y$ be an arbitrary element of $B$
Then, there exists a subset say $Y$ of set $B$ such that $y \in Y$.
Now $Y \subset B \Rightarrow y \in P(B) \Rightarrow y \in P(A)$
$[\because P(A)=P(B)]$
$\Rightarrow Y \subset A \Rightarrow y \in A$. Thus $y \in B \Rightarrow y \in A$
$\therefore B \subset A \quad \ldots (ii)$
From (i) and (ii), we get $A=B$