A subgroup of the central quotient and its orthogonal complement have orders multiplying to the order of the quotient

The group $V$ is elementary abelian of order $p^{2n}$, and $U$ is a subgroup of it, hence also a finite abelian $p$-group whose nonidentity elements have order $p$.

For each fixed $u\in V$ the map $v\mapsto b_z(v,u)$ is a homomorphism from $V$ to the additive group of ${\mathbb{F}}_p$, by additivity of $b_z$ in its first variable; $U^{\perp }$ is the intersection of the kernels of these maps as $u$ runs over $U$, hence a subgroup of $V$.

Choose a basis $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_m$ of $V$. Unique representation in a basis makes the assignment of coefficient families to elements a bijection, so $\mid U\mid =p^k$ and $p^m=\mid V\mid =p^{2n}$, whence $m=2n$.

Define $\mathrm{\Psi }:V\to V$ by $\mathrm{\Psi }(v)={\prod }_{i=1}^m{u}_i^{b_z(v,u_i)}$. It is a homomorphism, again by additivity in the first variable. If $\mathrm{\Psi }(v)$ is the identity then independence of the basis forces $b_z(v,u_i)=0$ for every $i$, and then additivity in the second variable gives $b_z(v,w)=0$ for every $w\in V$, so $v$ is the identity by triviality of the radical. Thus $\mathrm{\Psi }$ is injective, hence bijective because $V$ is finite.

Let $\pi :V\to U$ send ${\prod }_{i=1}^m{u}_i^{a_i}$ to ${\prod }_{i=1}^k{u}_i^{a_i}$; unique representation makes $\pi$ a well-defined homomorphism, and it is surjective onto $U$ because it fixes each $u_i$ with $i\le k$. The composite $\mathrm{\Phi }=\pi \circ \mathrm{\Psi }$ is therefore a surjective homomorphism from $V$ onto $U$, and $\mathrm{\Phi }(v)$ is the identity exactly when $b_z(v,u_i)=0$ for every $i\le k$, which by additivity in the second variable is exactly $v\in U^{\perp }$.

The first isomorphism theorem and Lagrange applied to $\mathrm{\Phi }$ give $\mid V\mid =\mid U^{\perp }\mid \mid U\mid$, that is $\mid U\mid \mid U^{\perp }\mid =p^{2n}$.