Let $A$ be a commutative ring with identity. If $A$ is a ring with only a finite set of prime ideals $p_1...p_n$ and moreover $\prod_{i=1}^n p_i^{k_i}=0$ for some k_i. Is $A$ then isomorphic to $\prod_{i=1}^nA_{(p_i)}$?

No. Let $A$ be a DVR. It has two prime ideals: the maximal ideal $p_1=\mathfrak m\subset A$ and $p_2=(0)\subset A$. So, $p_1p_2=0$, but $A$ is not a product (of two local rings). 

