Let $G$ be a finite abelian $p$-group (where $p$ is a prime). Suppose there exists a symmetric bilinear map $\delta\colon G\times G\to \mathbb{Q}/\mathbb{Z}$ such that the induced map $g\to\langle g,\;\rangle$, is an isomorphism from $G$ to $\mathrm{Hom} (G, \mathbb{Q}/\mathbb{Z})$.
Then, is it true that $G$ can be written as $\mathbf{Z}_p^n/\,\mathrm{im} (A)$ for some symmetric matrix (i.e. a linear map) $A\colon\mathbf{Z}_p^n\to\mathbf{Z}_p^n$ for some positive integer $n$? (Here $\mathbf{Z}_p$ denotes the $p$-adic integers)
EDIT2: As user74230 has pointed out, the isomorphism $G\to \mathbf{Z}_p^n/\,\mathrm{im} (A)$ has to respect the evident bilinear forms on both sides. (Otherwise, the answer is trivial and $A$ does not depend on $\delta$ as Amritanshu Prasad has pointed out).
EDIT1: I had this question while reading the following paper: http://arxiv.org/pdf/1402.5129v1. In Theorem 2 of page 4 it says that, given a $p$-group with duality pairing $(G,\delta )$ the probability (w.r.t. Haar measure) that $\mathrm{coker}(A)\cong (G,\delta)$ converges to Cohen-Lenstra type probability measure when $n\to\infty$. The existance of the duality pairing seems relevant for this theorem, but Amritanshu Prasad's comment suggests that $A$ can be chosen independent of the pairing which seems puzzling for me.