= Solution
Put $N=p^n$, $\alpha=|A|/N$, and use normalized <Fourier analysis on a finite abelian group>. If $F=1_A*1_A$ denotes unnormalized convolution and $h=N^{-1}F$, then
$$
\widehat h(\gamma)=\widehat{1_A}(\gamma)^2,
\qquad
\sum_\gamma|\widehat h(\gamma)|
=\sum_\gamma|\widehat{1_A}(\gamma)|^2
=\alpha
$$
by the <Parseval identity>.
Sample $k=O(m\varepsilon^{-2})$ characters independently, choosing $\gamma$ with probability $|\widehat h(\gamma)|/\alpha$, and attach the phase of $\widehat h(\gamma)$ to the sampled character. The <Marcinkiewicz–Zygmund inequality>, followed by averaging over $x$, shows that some sampled Fourier sum $g$ satisfies
$$
\|h-g\|_{L^{2m}(\mathbb E)}\leq\frac{\varepsilon\alpha}{2}.
$$
This is the sampling argument recorded in the <finite-field character approximation> principle.
Let $V$ be the intersection of the kernels of the sampled characters. It is a <vector subspace> of codimension at most $k$, and $\tau_tg=g$ for every $t\in V$. The <triangle inequality> therefore gives
$$
\|\tau_th-h\|_{L^{2m}(\mathbb E)}
\leq2\|h-g\|_{L^{2m}(\mathbb E)}
\leq\varepsilon\alpha.
$$
Returning from normalized convolution and normalized norm to $F$ and the counting norm multiplies the right side by $N^{1+1/(2m)}$. Hence
$$
\|\tau_tF-F\|_{2m}
\leq\varepsilon|A|N^{1/(2m)}
=\varepsilon|A|p^{n/(2m)}
$$
for every $t\in V$, proving the <finite-field convolution almost-periodicity theorem>.
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