When is every principal filter an intersection of ultrafilters? The question is in the title: what property does a lattice need to have such that for every element of the lattice $x$, there exists a set of ultrafilters in the lattice such that the intersection of that set of ultrafilters is identical to the principal filter of $x$?
The above is at least true for finite lattices. Is this property the same as being atomistic?
 A: Assuming that by "ultrafilter" you mean "maximal proper filter", this property of a bounded lattice $L$ is equivalent to the statement that for all $a,b\in L$ with $a\not\leq b$, there exists a nonzero $c\leq a$ such that $b\wedge c=0$.
Indeed, assuming that statement, then given a principal filter $F$ generated by an element $a$ and any $b\not\in F$, there is a nonzero $c\leq a$ such that $b\wedge c=0$.  Then an ultrafilter containing the filter generated by $c$ is an ultrafilter which contains $F$ but does not contain $b$.  So, $b$ is not in the intersection of all ultrafilters containing $F$.  Since $b\not\in F$ was arbitrary, this means $F$ is the intersection of all the ultrafilters that contain it.
Conversely, assuming your property, let $a,b\in L$ with $a\not\leq b$.  There must exist an ultrafilter $U$ containing the principal filter generated by $a$ such that $b\not\in U$.  Since $U$ is maximal, this means there exists $d\in U$ such that $b\wedge d=0$.  Then $c=d\wedge a$ is nonzero (since $a,d\in U$) and satisfies $c\leq a$ and $b\wedge c=0$.
This property holds for an atomistic lattice and the converse is true for finite lattices.  Indeed, if $L$ is atomistic and $a,b\in L$ satisfy $a\not\leq b$, then there exists some atom $c$ such that $c\leq a$ and $c\not\leq b$, and then $b\wedge c=0$.  Conversely, if $L$ is finite and satisfies your property, let $a\in L$ and let $b$ be the join of all the atoms below $a$.  Then if $a\not=b$, then $a\leq b$, so there exists a nonzero $c\leq a$ such that $b\wedge c=0$.  There then exists an atom $d\leq c$, and we have $d\leq a$ but $b\wedge d=0$.  This contradicts the definition of $b$.  Thus $a=b$, and $L$ is atomistic.
(In particular, this property does not hold in every finite lattice, since not every finite lattice is atomistic.  For instance, consider a chain with more than 2 elements.)
For infinite lattices, though, this condition is much weaker than being atomistic.  For instance, every Boolean algebra satisfies this condition, taking $c=a\wedge \neg b$.
