If $\alpha$ is a (limit) ordinal, then a subset $S\subseteq\alpha$ is club if $\alpha$ is closed as a subset of $\alpha$ under the order topology and unbounded in $\alpha$. The set of all sets containing a club forms a filter on the subsets of $\alpha$, called the club filter.
This definition can be extended in the following way. Let $C$ be any infinite set, and let $P$ be the set of all countable subsets of $C$. Then $S\subseteq P$ is club if $S$ is closed under unions of countable chains (closed), and for all $X\in P$, there is some $Y\in S$ with $X\subseteq Y$ (unbounded).
(One area where this notion of clubness appears is in proofs of a Lowenheim-Skolem theorem for infinitary logic; see e.g. http://www.math.uic.edu/~marker/dwk.pdf.)
Given a set $C$, we can take the set of all subsets of $P$ which contain a club (in the generalized sense); call this the club filter on $P$.
My question is: when is the club filter in this latter case an ultrafilter?
If I understand things correctly, there should be no possibility of the club filter being an ultrafilter if we assume AC. However, in the original sense of the word club, the club filter on $\omega_1$ is an ultrafilter assuming AD, so this leads me to believe that, in ZF + AD, there might be interesting sets $C$ the club filter of which is an ultrafilter.
In particular, what kinds of choice need to fail at $C$ or $P$ in order for the club filter on $P$ to be an ultrafilter?
I hope this question is meaningful; I don't have much background knowledge of models of set theory in which choice fails.