Let $p$ be a prime and $G$ be a nonabelian group of order $p^4$. Let $H$ be a subgroup of $G$ maximal with respect to $H$ being normal in $G$ and $H$ being abelian. We have to show that $|H|=p^3$.
I want the following approach.
Since $G$ is a nonabelian $p$-group, $\{1\}\neq Z(G) \underset{\neq}{\triangleleft} G$ , hence by maximality of $H$, $|H|\neq 1$.
Now suppose $|H|=p$. By NC Lemma we obtain, $\frac{G}{C_G(H)} \hookrightarrow Aut(H)$. Since $|Aut(H)|=p-1$, only possibility is $C_G(H)=G$, then $H \subset Z(G)$. By maximality of $H$, we must have $H=Z(G)$. I stuck here. I thought about taking the $p$-group $G/H$. Then $Z(G/H)=K/H$, where $H \lneq K \unlhd G$. Further I can have $[K,G]=H$. How can I complete from here?
Next situation is that $|H|=p^2$. Then I can only found by NC lemma $|C_G(H)|\neq p^2$, since $|Aut(H)|=p(p-1)$ or $(p^2-1)(p^2-p)$. Then I cannot complete as well.
I will be really grateful and thankful if you help me to complete it the way I have approached.
Case $|H|=p$.
Look at the possible orders of the second centre $K$.
(i) If $|K|=p^2$ then it is a normal abelian subgroup properly containing $H$.
(ii) $|K|=p^3$ cannot occur, since in that case the quotient $(G/Z(G))/Z(G/Z(G))$ would be cyclic, and so $G/Z(G)$ would be abelian and we would have $|K|=p^4$.
(iii) If $|K|=p^4$ then we have that $G/H$ is abelian. If we take any $x\not\in H$ of order $p$ modulo $H$ then $\langle H,x\rangle/H$ is a normal subgroup of $G/H$, and hence $\langle H,x\rangle$ is a normal subgroup of $G$. It is also abelian as $H$ is the centre. So this case cannot arise.
Case $|H|=p^2$.
As you point out, $N(H)/C(H)$ has order $1$ or $p$. As $H$ is normal, $N(H)=G$.
(i) If $N(H)/C(H)$ has order $1$, then $H$ is the centre of $G$. As $G/H$ has order $p^2$ it is abelian. So let $x\not\in H$ and argue as above that $\langle H,x\rangle$ is normal and abelian, a contradiction.
(ii) Suppose then that $N(H)/C(H)$ has order $p$. Then $C(H)$ is a normal subgroup of $G$, and contains the subgroup $H$ (since $H$ is abelian). So $H$ is central in $C(H)$ and the quotient is cyclic: so $C(H)$ is abelian. So this case cannot arise.