= Solution
Induct on the order of the <p-group>. Every complex <irreducible representation> of an abelian group is one-dimensional: commuting group operators preserve a common eigenspace, and irreducibility makes that eigenspace the whole space. This handles the abelian cases.
If $V$ is not faithful, it is an irreducible representation of a smaller quotient $G/K$, where $K$ is its kernel. Induction expresses it as induced from a one-dimensional representation of a subgroup of $G/K$. Pulling back the subgroup and inflating the character gives the same induction statement for $G$, because the coset spaces of the quotient subgroup and its inverse image agree.
Now suppose $V$ is faithful and $G$ is nonabelian. Use the allowed abelian normal subgroup $A$ not contained in the centre. If restriction to $A$ were isotypic, its irreducible type would be one-dimensional, so every $a\in A$ would act as a scalar. It would therefore commute with every representing matrix of $G$. Faithfulness would imply $ga=ag$ in $G$ for every $g$, contradicting the choice of $A$. By part (a), $V=\operatorname{Ind}_H^GW$ for an irreducible representation of a proper subgroup $H$. Induction on $H$ gives $W=\operatorname{Ind}_K^H\chi$ with $\chi$ one-dimensional. Transitivity of induction yields
$$
\boxed{V=\operatorname{Ind}_K^G\chi.}
$$
This proves the <monomiality of irreducible representations of finite p-groups>.
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