This is not a complete answer, but it's too long to be a comment.

I think I found an example for which the bound $\lceil \log_2 N_x \rceil$ is not tight.

Consider the following poset. The ground set is $\{a_1, a_2, b_1, b_2\}$, and $a_i$ is smaller than $b_j$ for all $i,j\in\{1,2\}$. The other pairs are incomparable. (The Hasse diagram is a $4$-cycle).

Let me identify the monotone properties with the upsets of the poset. This poset has seven upsets: $\emptyset$, $\{b_1\}$, $\{b_2\}$, $\{b_1,b_2\}$, $\{a_1,b_1,b_2\}$, $\{a_2,b_1,b_2\}$, $\{a_1,a_2,b_1,b_2\}$, and this poset has seven antichains since the antichains are in one-to-one correspondence with the upsets. So, $\lceil \log_2 N_x \rceil=\lceil \log_2 7 \rceil = 3$ for this poset.

Now, by adversary argument I'll show that any strategy needs at least four queries (so needs to query all elements). Let's fix an arbitrary strategy.

If the strategy first queries $a_1$, then the adversary answers "$P(a_1)$ doesn't hold." Then, we are left with five possibilities: $\emptyset$, $\{b_1\}$, $\{b_2\}$, $\{b_1,b_2\}$, $\{a_2,b_1,b_2\}$. Thus, to determine which is the case, we need at least $\lceil \log_2 5\rceil = 3$ more queries. In total, we need four queries. The same argument applies if the first query is $a_2$.

If the strategy first queries $b_1$, then the adversary answers "$P(b_1)$ holds." Then, we are left with five possibilities: $\{b_1\}$, $\{b_1,b_2\}$, $\{a_1,b_1,b_2\}$, $\{a_2,b_1,b_2\}$, $\{a_1,a_2,b_1,b_2\}$. Therefore, we need at least three more queries as before. In total, we need four queries. The same argument applies when the first query is $b_2$.

If we take $k$ parallel copies of this poset, then it has $7^k$ antichains, and thus the proposed bound is $\lceil \log_2 7^k \rceil = 3k$. But, since each of the copies needs four queries, we need at least $4k$ queries.

Probably, there is a larger poset with larger gap. But this argument can only improve the coefficient.

Here, the problem looks to be a situation where no query partitions the search space evenly. In such a case, the adversary can force the larger half to remain.

This is not a complete answer, but it's too long to be a comment.

I think I found an example for which the bound $\lceil \log_2 N_X \rceil$ is not tight.

Consider the following poset. The ground set is $X=\{a_1, a_2, b_1, b_2\}$, and $a_i$ is smaller than $b_j$ for all $i,j\in\{1,2\}$. The other pairs are incomparable. (The Hasse diagram is a $4$-cycle).

Let me identify the monotone properties with the upsets of the poset. This poset has seven upsets: $\emptyset$, $\{b_1\}$, $\{b_2\}$, $\{b_1,b_2\}$, $\{a_1,b_1,b_2\}$, $\{a_2,b_1,b_2\}$, $\{a_1,a_2,b_1,b_2\}$, and this poset has seven antichains since the antichains are in one-to-one correspondence with the upsets. So, $\lceil \log_2 N_X \rceil=\lceil \log_2 7 \rceil = 3$ for this poset.

Now, by adversary argument I'll show that any strategy needs at least four queries (so needs to query all elements). Let's fix an arbitrary strategy.

If the strategy first queries $a_1$, then the adversary answers "$P(a_1)$ doesn't hold." Then, we are left with five possibilities: $\emptyset$, $\{b_1\}$, $\{b_2\}$, $\{b_1,b_2\}$, $\{a_2,b_1,b_2\}$. Thus, to determine which is the case, we need at least $\lceil \log_2 5\rceil = 3$ more queries. In total, we need four queries. The same argument applies if the first query is $a_2$.

If the strategy first queries $b_1$, then the adversary answers "$P(b_1)$ holds." Then, we are left with five possibilities: $\{b_1\}$, $\{b_1,b_2\}$, $\{a_1,b_1,b_2\}$, $\{a_2,b_1,b_2\}$, $\{a_1,a_2,b_1,b_2\}$. Therefore, we need at least three more queries as before. In total, we need four queries. The same argument applies when the first query is $b_2$.

If we take $k$ parallel copies of this poset, then it has $7^k$ antichains, and thus the proposed bound is $\lceil \log_2 7^k \rceil = 3k$. But, since each of the copies needs four queries, we need at least $4k$ queries.

Probably, there is a larger poset with larger gap. But this argument can only improve the coefficient.

Here, the problem looks to be a situation where no query partitions the search space evenly. In such a case, the adversary can force the larger half to remain.