= Solution
For every <set> $S$, the definition of a <power set> gives
$$
\begin{aligned}
S\in\mathcal P(A\cap B)
&\Longleftrightarrow S\subseteq A\cap B\\
&\Longleftrightarrow (S\subseteq A\text{ and }S\subseteq B)\\
&\Longleftrightarrow S\in\mathcal P(A)\cap\mathcal P(B).
\end{aligned}
$$
Equality by identical membership therefore proves
$$
\boxed{\mathcal P(A\cap B)=\mathcal P(A)\cap\mathcal P(B).}
$$
The essential observation is that being a <subset> of a <set intersection> means being a <subset> of both factors.
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