Binary search generalizations for posets?
I have not thought this through very much, so please correct me if I am wrong.

Say $w$ is the width of the poset.


For the poset which is the union of $w$ disjoint chains you need at least $w\log n$ evaluations of $P$ by just applying the standard lower bound on the query complexity of binary search to each chain.
Since you give comparisons for free, you can compute a chain decomposition of the poset into $w$ chains for free. Do binary search on each chain to identify the first element that satisfies $P$. Then go over the identified elements and remove any dominated ones. Number of evaluations of $P$ is $O(w\log n)$. This identifies all maximal elements, as there can be at most one maximal element per chain.




ADDED: Actually I am seeing a simple recursive algorithm to do much better ($O(n)$) for the lattice of subsets $2^{[n]}$ (EDIT: domotor described the general strategy in his answer). Here I am assuming $P$ is monotonic downwards (i.e. the subsets $\{X: P(X) = 1\}$ form a lower set), which is I think what you mean. So, here is the algorithm to find a member of the lower set:

a) Test $P(\emptyset)$. If 0, then stop.

b) Test $P(\{n\})$.  

b.i) If 0, then recurse on $2^{[n-1]}$ (OK, since no set containing $n$ can be in the lower set). 

b.ii) If 1, then there exists a member of the lower set in the sublattice $\{X: n \in X\}$. This sublattice is isomorphic to $2^{[n-1]}$ so once again we can recurse. More precisely, we can run the algorithm for $2^{[n-1]}$, but when the algorithm asks to evaluate $P(Y)$, we evaluate $P(X)$ where $X = Y \cup \{n\}$.

So in each step we recurse on a sublattice which is half the size of original one. Overall, we need to evaluate $P$ at most $2n$ times (in fact you can implement the algorithm to evaluate the predicate $n+1$ times as Yoshio points out, since you only need to check $\emptyset$ once).