Any abelian normal subgroup normalizes an abelian subgroup of maximum order

Existential version
Any abelian normal subgroup of a group of prime power order (i.e., any  abelian normal subgroup of group of prime power order)  normalizes an  abelian subgroup of maximum order.

In symbols: let $$P$$ be a group of prime power order and $$B$$ be an abelian normal subgroup of $$P$$. Then, there exists an abelian subgroup of maximum order $$A$$ in $$P$$ such that $$B \le N_P(A)$$.

Replacement version
let $$P$$ be a group of prime power order and $$B$$ be an abelian normal subgroup of $$P$$. Let $$A$$ be any abelian subgroup of maximum order in $$P$$. Then, there exists an abelian subgroup of maximum order $$A_\infty$$ in $$P$$ such that $$B \le N_P(A_\infty)$$ and such that $$A_\infty$$ is contained in the subgroup $$\langle A^B \rangle$$, i.e., the closure of $$A$$ under the action of $$B$$ by conjugation.

Related facts
Note that this fact is uninteresting for small orders for the following silly reason: for small orders, it is also true that, among the abelian subgroups of maximum order, there exists a normal subgroup. The existence of a normal subgroup that is abelian of maximum order is obviously substantially stronger than this statement.

Below, we indicate which of a bunch of stronger statements is true, and why. In particular, we note from this that the smallest examples of interest where the stronger statements do not hold are $$2^9 = 512$$ for the prime 2 and $$p^{10}$$ (or possibly higher for some $$p$$, depending on other replacement results that are dependent on the value of $$p$$ -- see Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one for instance) for odd primes.

Facts used

 * 1) uses::Stable version of Thompson's replacement theorem for abelian subgroups, which in turn follows from uses::Thompson's replacement theorem for abelian subgroups

Proof of replacement version
The proof follows directly from Fact (1), and the observation that the hypotheses apply because, on account of $$B$$ being abelian normal, any abelian subgroup of maximum order must normalize it.

Proof of existential version
This is weaker than, and hence follows from, the replacement version.