Yes. Note that trivially $|P|\leq|\mathcal{P}(X)|$ since $P\subseteq\mathcal{P}(X)$. If $|P|=|\mathcal{P}(X)|$, then we can define a surjection $X\to\mathcal{P}(X)$ by mapping each element of $X$ to the element of $P$ that contains it and then composing with a bijection $P\to\mathcal{P}(X)$. This is impossible by Cantor's theorem.