= Solution
<Conjugation> by $P$ partitions $H$ into <orbits>. Each non-singleton <orbit> has size divisible by $p$, by the <orbit-stabilizer theorem>. The singleton <orbits> are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv0\pmod p$. This <set> contains $1$, so $\boxed{|H\cap Z(P)|\geq p}$.
For the deduction we also give the representation-theoretic ingredient behind <Burnside's theorem>: a nonabelian finite <simple group> has no nontrivial <conjugacy class> of <prime power> size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne1$ has class size $p^b$. For an <irreducible character> $\chi$ of degree $d$ coprime to $p$, the <class sum> acts as the scalar $p^b\chi(g)/d$, which is an <algebraic integer>: the <class sum> has an integer-entry <matrix> in the <regular representation>, which contains every <irreducible representation>, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi(g)$ is an <algebraic integer>. Bezout's identity then makes $\chi(g)/d$ an <algebraic integer>. All its <algebraic conjugates> have modulus at most $1$, because <character values of a representation> are sums of $d$ <roots of unity>. An <algebraic integer> with this property is zero or a <root of unity>: the monic <polynomials> of all its powers have uniformly bounded <integer> coefficients, so two powers must coincide unless it is zero.
Every nontrivial <irreducible representation> of a <simple group> is faithful. If $\chi(g)/d$ were a <root of unity>, equality in the triangle inequality would force its representing <matrix> to be scalar, hence $g$ central, impossible. Thus $\chi(g)=0$ whenever $p\nmid d$, except for the trivial <character of a representation>. Orthogonality of the identity column and the $g$ column in the <character table> now gives
$$
0=1+\sum_{p\mid\chi(1)}\chi(1)\overline{\chi(g)}.
$$
This would make $1/p$ an <algebraic integer>, contradicting the fact that rational <algebraic integers> are <integers>. The same proof covers class size $1$ directly through triviality of the center.
If an abelian <subgroup> $A$ had index $p^a$, every $g\in A\setminus\{1\}$ would have $A\subseteq C_G(g)$, making its class size a <divisor> of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial <p-group>, whose center is nontrivial by the first argument with $H=P$. Therefore \b[a nonabelian <simple group> has no abelian <subgroup> of <prime power> index].
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