Let $\mathbb{M}$ be a countable transitive model of ZFC.
I understand that $\omega^\mathbb{M} = \omega$ but $\mathcal{P}(\omega)^\mathbb{M} \neq \mathcal{P}(\omega)$, for $\mathcal{P}(\omega)^\mathbb{M}$ is countable from an external point of view. So $\mathcal{P}(\omega)^\mathbb{M}$ is the set of all the subsets of $\omega$ that belongs to $\mathbb{M}$.
Here is where I get confused : how is it possible that some subset of $\omega$ do not belong to $\mathbb{M}$ ? After all, all the finite ordinals belong to $\mathbb{M}$ and $\mathbb{M}$ is a model of ZFC. Why is it that with the same ingredients (the finite ordinals) and with the same construction rules (the axioms of ZFC), we manage to build all the subsets of $\omega$ in $V$ but we are not able to do so in $\mathbb{M}$ ?
The key point here, I think, is that we don't build the sets. The universe is given to us, it is just there. Of course, we are able to prove that by iterating transfinitely the power set operation from $\varnothing$, we can end up with any given set. But the power set operation is in some sense atomic to the universe.
And this is the case when $V$ itself is not $L$, and we can define a subclass of $V$ which is also a model of $\sf ZFC$, but doesn't agree with $V$ on the power set of at least one ordinal (sometimes $\omega$, sometimes not, it doesn't matter). In fact, that is an even better question, suppose that $V=L[x]$ for some real number $x$. Then $V$ and $L$ disagree on $\mathcal P(\omega)$, even though both contain all the ordinals.