# if A is in P and B is NP, is A ≤p B?

if A is in P and B is NP, then is A polynomial time reducible to B?

Could anyone prove a prove or disprove for it?

If $$B$$ is neither empty nor the universal set, then the reduction can be performed and is trivial.
You need to have two strings $$w_1 \in B$$ and $$w_2 \notin B$$. In the reduction of membership problem $$(A,x)$$, we trivially solve the membership in $$A$$ in polynomial time: if $$x \in A$$ then return $$w_1$$ else $$w_2$$.
Assuming you're talking about polynomial-time many-one reductions (i.e., Karp reductions), the claim is false. Consider the binary alphabet, let $$A = \{0,1\}^*$$ and $$B=\emptyset$$. Clearly $$A$$ is in $$\mathsf{P}$$ and $$B$$ is in $$\mathsf{P} \subseteq \mathsf{NP}$$, yet $$A$$ is not Karp-reducible to $$B$$.