Let $G$ be a group of order $p^4$ for some prime number $p$. Suppose $G$ has a normal subgroup $H$ of order $p^2$. Then $G$ has an abelian subgroup $K$ with $|K|\geq p^3$.
First, by acting $G$ onto $H$ by conjugation, I have an injection $G/C_G(H)\hookrightarrow \mbox{Aut}(H)$. Since $H$ is group of order $p^2$, I checked that whether $p$ is even or odd primes, $p\mid |\mbox{Aut}(H)|,p^2\nmid |\mbox{Aut}(H)|$. Since $G/C_G(H)$ is $p$-group, I concluded that $G/C_G(H) = 1$ or $\simeq \Bbb Z/p$. Now I want to say that $C_G(H)$ is abelian but stuck. Please help.
Every group $T$ such that $T/Z(T)$ is cyclic, is Abelian (exercise). Now apply it to $C_G(H)$. This gives you what you want in case $C_G(H)/H$ is cyclic of order $p$. Now from what you proved, since $G/H$ is Abelian of order $p^2$ it handles the case when $G/C_G(H)$ is cyclic of order $p$. If this factor-group is trivial, then take any coset $aC_G(H)\in C_G(H)/H$ of order $p$ (it exists by Cauchy's theorem because in this case $C_G(H)/H$ is of order $p^2$). Since $a$ centralizes $H$, $\langle H,a \rangle$ is Abelian of order $p^3$.