On page 52 of this paper, Iwasawa considered the bilinear symmetric non-degenerate pairing $\Phi_n \times \Phi_n \rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ defined by $$\langle \alpha, \beta \rangle_n := \text{the class of } T_n(\alpha\beta) \text{ in } \mathbb{Q}_p / \mathbb{Z}_p$$ where $\Phi_n = \mathbb{Q}_p(\zeta_n)$, $\zeta_n$ being $p^{n+1}$-th root of unity and $T_n$ is the trace of the field extension $\Phi_n/\mathbb{Q}_p$. He then wrote

For any closed subgroup $A$ of [the additive group] $\Phi_n$, we denote by $A^\perp$ the annihilator of $A$ in $\Phi_n$ relative to this pairing. Then $A^\perp$ is a closed subgroup of $\Phi_n$ such that $(A^\perp)^\perp = A$, ...

Unfortunately, I do not see the reasons for his statement that $(A^\perp)^\perp = A$. I know that it is true in the case of non-degenerate bilinear form and $A$ is a *subspace* but the pairing given is not a bilinear form. So how can I prove that?