My question: Is the bilinear form $B$ defined below non-degenerate in general?
Let $P$ be a non-abelian special $p$-group $\left(Z(P) = [P,P] = \Phi(P)\right)$ with $Z = Z(P) \cong C_{p}^{n}$ ($n \geq 1$). Let $z_{1},z_{2},\dots,z_{n}$ be a minimal generating set for $Z$. Define $B:P/Z \times P/Z \rightarrow \mathbb{F}_{p}$ by $$ B(\overline{x},\overline{y}) = a \,\, \text{where} \,\, [x,y] = \prod_{i = 1}^{n}z_{i}^{a_{i}} \,\, \text{and} \,\, \sum_{i = 1}^{n} a_{i} \equiv a \,\, \text{mod} \,\, p. $$ Then $B$ is well defined since $[xz,yz'] = [x,y]$ for all $z,z' \in Z$.
Since $[x,y] \in [P,P] = Z(P)$, we have $$ [xx',y] = [x,y]^{x'}[x',y] = [x,y][x',y] $$ and $$ [x,yy'] = [x,y'][x,y]^{y'} = [x,y'][x,y] = [x,y][x,y']. $$ Therefore, $B(\overline{x}\overline{x}',\overline{y}) = B(\overline{x},\overline{y}) + B(\overline{x}',\overline{y})$ and $B(\overline{x},\overline{y}\overline{y}') = B(\overline{x},\overline{y}) + B(\overline{x},\overline{y}')$. Furthermore, by what was just shown, for $\lambda \in \mathbb{F}_{p}$ we have $$ B(\overline{x}^{\lambda},\overline{y}) = B(\overline{x}\overline{x}\cdots \overline{x},\overline{y}) = \lambda B(\overline{x},\overline{y}) $$ and $$ B(\overline{x},\overline{y}^{\lambda}) = B(\overline{x},\overline{y}\overline{y}\cdots \overline{y}) = \lambda B(\overline{x},\overline{y}) $$ Hence $B$ is bilinear.
If $n = 1$, then it is easy to see that $B$ is non-degenerate, but is $B$ non-degenerate in general?