I'm trying to see if $\operatorname{Hom}_{R/\mathfrak{p}}(\frac{\mathfrak{p}^{n}}{\mathfrak{p}^{n+1}},X)\cong \operatorname{Hom}_{R}(\frac{\mathfrak{p}^{n}}{\mathfrak{p}^{n+1}},X)$, for $R$ any commutative unitarian ring, $\mathfrak{p}$ a prime ideal of $R$ and $X$ a $R$-module such that $\operatorname{Ass}(X)=\left\lbrace \mathfrak{p}\right\rbrace$.
I think that this is true. Take the arrow $\operatorname{Hom}_{R}(\frac{\mathfrak{p}^{n}}{\mathfrak{p}^{n+1}},X)\rightarrow \operatorname{Hom}_{\frac{R}{\mathfrak{p}}}(\frac{\mathfrak{p}^{n}}{\mathfrak{p}^{n+1}},X)$ that sends any $R$-homomorphism to itself as a $R/\mathfrak{p}$-homomorphism. This should be surjective, but I'm not sure whether it is injective.
Am I correct? If I'm not, can you think of a counterexample? Is the $\operatorname{Ass}(X)=\left\lbrace \mathfrak{p}\right\rbrace$ hypothesis really needed?
You are correct.
If $R$ is a ring and $I$ is an ideal, then for any $R/I$-modules $M$ and $N$, there is an isomorphism $\operatorname{Hom}_R(M,N) \cong \operatorname{Hom}_{R/I}(M,N)$. Both of these are subsets of the set of maps from $M$ to $N$ as sets. What you need to check is that a map $f:M\to N$ is an $R$-homomorphism if and only if it is an $R/I$-homomorphism.
The hypothesis is not really needed: all you need in your case is for $X$ to be a $R/\mathfrak p$-module, for the question to even make sense.