It is equivalent to AC.
Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial function on $A$ for which $p(a)\in a$ for every $a\in\text{dom}(p)$. We place the forcing order on $\P$, so that $p\leq q$ if $p$ extends $q$ to a larger domain, or equivalently, $q=p\upharpoonright\text{dom}(q)$. (In particular, being lower in the order means having more information, larger domain etc.) The empty function is the largest element of $\P$. Let us also add an object $\bot$ to $\P$ below all others.
The basic motivating idea is that $\P$ is the forcing to add a choice function for $A$, augmented with $\bot$.
This is a lattice, because any two partial functions $p$, $q$ have a least upper bound $p\vee q$, which is their common part as functions, and a greatest lower bound, which is their union $p\cup q$ if they are compatible as functions, and otherwise $\bot$.
I assume that ultrafilters for you cannot be the whole lattice (since otherwise the ultrafilter assertion would be trivialized). Every proper filter in $\P$, I claim, corresponds to a partial choice function, since it cannot contain $\bot$ and so all elements of it must be compatible as functions. Furthermore, an ultrafilter must be totally defined, since otherwise we could extend it by defining the choice function on one more set $a\in A$.
So from an ultrafilter in $\P$ we get a choice function on $A$.