The problem is as follows:
Let $I\subseteq J$ be ideals in a Noetherian ring. Show that if $I_{p}=J_{p}$ for every associated prime $p$ of $I$,then $I=J$.
It seems reasonable to consider $J/I\subseteq R/I$ but I couldn't go on.
Let M be an $R-\mathrm{module}$. A prime ideal $p$ is an associated prime of $M$ if there exists nonzero $x\in M$ such that $p=\mathrm{ann}_{R}(x)$.
Hint: We have that $\operatorname{Ass}(J/I) \subset \operatorname{Ass}(R/I)$ (why?). Now, use this together with the hypothesis $I_P = J_P, \, \forall P \in \operatorname{Ass}(R/I)$ to conclude that $J/I = 0$. In doing so, recall that for a finite $R$-module $M$ it is always the case that $\operatorname{Ass}(M) \subset \operatorname{Supp}(M)$.
Remark: The definition of associated prime that you gave is correct, but recall that it is a convention that when one talks about associated primes of an ideal $I$, they really mean the associated primes of the quotient $R/I$. As Mohan says, this can be quite confusing and takes some time to get used to.