= Solution
<Frobenius reciprocity> is the adjointness of <character> induction and restriction: for <class functions> $f$ on $H$ and $u$ on $G$,
$$
\boxed{\langle f^G,u\rangle_G=\langle f,u_H\rangle_H}.
$$
For <characters> this says that the multiplicity of an irreducible $G$-character in an <induced character> equals the corresponding multiplicity after restriction. To prove the identity directly, insert the induced <class function> formula and interchange the two finite sums. For each $x$, substitute $g=xhx^{-1}$ with $h\in H$. Since $u$ is constant on <conjugacy classes>,
$$
\begin{aligned}
\langle f^G,u\rangle_G
&=\frac1{|G||H|}\sum_{x\in G}\sum_{h\in H}f(h)\overline{u(xhx^{-1})}\\
&=\frac1{|H|}\sum_{h\in H}f(h)\overline{u(h)}
=\langle f,u_H\rangle_H.
\end{aligned}
$$
This proves the theorem for all <class functions>, not just <irreducible characters>.
For the group-theoretic conclusion, suppose that $p$ divides $|G'\cap Z(G)|$. A central element of order $p$ lies in every <Sylow p-subgroup>: adjoining it to a <Sylow subgroup> gives a p-subgroup, so maximality forces it into that <subgroup>. Hence there is a <subgroup> $Q\leq P\cap G'\cap Z(G)$ of order $p$. Choose a nontrivial <linear character> $\lambda$ of $Q$.
Since $P$ is abelian, $\lambda$ extends to a <linear character> $\mu$ of $P$. One direct justification is to extend from an abelian <subgroup> $A$ to $\langle A,x\rangle$ one generator at a time: if $d$ is the least positive exponent with $x^d\in A$, choose a complex $d$th root of $\lambda(x^d)$ as the value at $x$. The formula on $ax^j$ is then well defined and multiplicative. Iteration yields the claimed extension to $P$.
The <induced character> $\mu^G$ has degree $[G:P]$, which is prime to $p$. In its decomposition into <irreducible characters>, at least one constituent $\chi$ must have degree prime to $p$; otherwise the total degree would be divisible by $p$. <Frobenius reciprocity> also gives $\langle\chi_P,\mu\rangle_P>0$ for this constituent. Restriction to the central <subgroup> $Q$ forces
$$
\chi_Q=\chi(1)\lambda.
$$
Indeed, the <induced representation> itself restricts to $Q$ as copies of $\lambda$, because conjugation by every coset representative is trivial on $Q$; alternatively this follows from <Schur's lemma> and the positive multiplicity after restriction.
Each $q\in Q$ consequently acts in the <representation> affording $\chi$ as the scalar $\lambda(q)$. Taking its <determinant character> gives
$$
(\det\chi)_Q=\lambda^{\chi(1)}.
$$
The left side is trivial since $Q\leq G'$ and every <linear character> kills $G'$. The right side is nontrivial since $\lambda$ has order $p$ and $p\nmid\chi(1)$. This contradiction proves
$$
\boxed{p\nmid|G'\cap Z(G)|}.
$$
The argument is the <central commutator has no torsion from an abelian Sylow subgroup> determinant obstruction.
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