$\mathsf{ACF}$ is the proposition that every set of nonempty finite sets has a choice function. It can be shown that $\mathsf{BPIT} \Rightarrow \mathsf{ACF}$, because $\mathsf{BPIT}$ implies that every set can be linearly ordered, so we just linearly order the union of all of the finite sets, and take the least element. I'm interested in whether the two are equivalent.
My intuition says that the $\textsf{ACF} \Rightarrow \textsf{BPIT}$ implication should hold. Let $\mathcal{F}$ be a filter on $X$. My idea is to take each pair of the form $\{S,\, X \setminus S\}$, and use $\mathsf{ACF}$ to choose a single element of each one, and to somehow <wave hands> use this to build up the ultrafilter, </wave hands> which is where I get stuck.
ACF is implied by the ordering principle (any set can be linearly ordered), which is strictly weaker than BPIT/Ultrafilter Lemma.
See Jech's "Axiom of Choice" or the wikipedia article on "linear extension" for more details.
EDIT: Upon further review, Jech (in section 7.3 of the above book) also provides a result that ACF is strictly weaker than the ordering principle. The proofs I'm aware of for the independence of OP from ACF, and of BPIT from OP, involve some voodoo (i.e. set theoretic tomfoolery I don't understand at the moment) with permutation models, so I can't really be more informative there.