A filter $U$ over a boolean algebra $A$ (isomorphic to a powerset algebra) "preserves" a join $a = \bigcup_{i\in I}a_i$, if $a\in U$ implies $a_i\in U$ for some $i\in I$. A join $a$ is infinite if $I$ is. There exist ultrafilters preserving countable sets of infinite joins and, moreover, for an arbitrary non-zero element $e \in A$, there is such an ultrafilter containing $e$.
The question is: if we have given a subset $S\subset A$ with finite intersection property (each non-empty finite subset $T\subseteq S$ has a non-zero meet), and a countable set of infinite joins in $A$, does there exist an ultrafilter containing $S$ and preserving these joins? If the general answer is negative, are there any additional conditions on $A$ or $S$ that ensure the existence of such an ultrafilter?
This is certainly not true in general. For instance, if you take a single infinite join $1=\bigcup a_i$ where the $a_i$ are disjoint, and let $S$ consist of the complements of all the $a_i$, then no proper filter containing $S$ can preserve this join. More generally, an obvious necessary condition is that for each join $a=\bigcup a_i$ which you want to preserve, the filter generated by $S$ cannot contain every $a\setminus a_i$. (Conversely, it is similarly obviously both necessary and sufficient that it be possible simultaneously to choose an $i$ for each join such that $S$ together with all the elements $\neg(a\setminus a_i)$ still has the finite intersection problem. I don't see any way to simplify that condition in general, though.)