This is basically the same answer as Asaf's, but since I've added a little more details, I guess there is not much harm in posting it. (I did not want to throw it away, when the post was already prepared.)
If $f$ is order isomorphism, then it must map atoms of one ordered set to atoms of the other one.
In this case it implies that singletons are maps to singletons, i.e. $f(\{x\})$ is a singleton for every $x\in A$. Thus the only possibility for the function $F$ is $F(x)=y \Leftrightarrow f(\{x\})=\{y\}.$
Since $\mathcal P(X)$ is an atomic lattice, every element of $\mathcal P(X)$ is uniquely determined by atoms which are below it. This shows that images of singletons uniquely determine an order-isomorphism. (And so every order isomorphism is indeed of this form.)