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, $q=p\upharpoonright\text{dom}(q)$. The empty function is the largest element of $\P$. Let us also add an object $\bot$ to $\P$ below all others.
This is a lattice, because any two partial functions $p$, $q$ have a least upper bound $p\vee q$, which is their common part, and a greatest lower bound, which is $p\cup q$ 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 all elements of it must be compatible, and furthermore, an ultrafilter must be totally defined, since otherwise we could define it at one more point.
So from an ultrafilter in $\P$ we get a choice function on $A$.