To prove that $P(A \cap B) = P(A) \cap P(B)$, one must demonstrate that each element of $P(A \cap B)$ is contained in $P(A) \cap P(B)$ and vice versa. This involves using the definitions of power sets and intersections. The proof starts by assuming an element $X$ belongs to $P(A \cap B)$, which implies $X \subseteq A \cap B$. From the properties of intersections, it follows that $X$ must also be a subset of both $A$ and $B$, thus belonging to $P(A)$ and $P(B)$. Ultimately, the equality holds as both sets contain the same elements.