$\wp^{(\omega)}(A)=?$ Let $A$ be a set,
$$\wp^{(0)}(A)=A$$
$$\wp^{(n+1)}(A)=\wp(\wp^{(n)}(A))$$
But what sense does $\wp^{(\alpha)}(A)$ make where $\alpha$ is a limit ordinal number?
The most natural way is let
$$\wp^{(\alpha)}(A)=\lim_{\xi \uparrow \alpha}\wp^{(\xi)}(A),$$ but what is this 'limit' means? Note that $\wp^{(n)}(A)$ probably not the subset of $\wp^{(n+1)}(A)$.  
 A: If you’re going to define it at all, the only reasonable definition is that $$\wp^{(\eta)}(A)=\bigcup_{\xi<\eta}\wp^{(\xi)}(A)$$ when $\eta$ is a limit ordinal.
For $n\in\omega$, elements of $\wp^{(n)}(A)$ are (intuitively speaking) sets that have members of $A$ buried inside $n$ layers of curly braces. You can’t have things buried exactly $\omega$ layers deep, because you need an outer layer of braces, but you can have things buried at most $\omega$ layers deep, though of course that just turns out to be things buried less than $\omega$ layers deep.
It’s more natural to look at the operation $A\mapsto A\cup\wp(A)$: then it’s clear that at limit stages you just want to take the union of the earlier stages.
A: Another possibility would be to take $\mathscr P^{(\alpha)}(A)$ to be the set of sequences $(X_\beta)_{\beta<\alpha}$ such that $X_\beta \subseteq \mathscr P^{(\beta)}(A)$ for all $\beta$ and $X_\beta = \bigcup X_{\beta+1}$ whenever $\beta+1<\alpha$. That is isomorphic to the usual iterated power set when $\alpha$ is finite, but generalizes more smoothly to $\omega$.
(It looks like it will need a bit more work to stay smooth above $\omega$, though).
(In fact it needs a bit more work in order to work at all, since as it stands it doesn't depend on $A$ at all).
