Suppose $S$ and $T$ are sets such that $|S|=|T|$ Prove that $|\mathcal{P}(S)|=|\mathcal{P}(T)|$.
To start with, $|S|<|\mathcal{P}(S)|$; $|T|<|\mathcal{P}(T)|$.
Just the statement itself sounds so obvious, that it confuses me a lot. I guess it might have something to do with Cantor-Bernstein Theorem but I do not know how
No. You don't need the Cantor-Bernstein theorem.
Since $|S|=|T|$, you have a bijection $f$ between $S$ and $T$. Define $\varphi:P(S)\to P(T)$ like as \[\varphi(X) = \{ f(x) : x\in X\}.\] You can check that $\varphi$ is 1-1 and onto.