To sum up, the result is false in general (see first part of the post below), trivially false for nonincreasing sequences $(a_n)$ if $p<2$ (consider $a_n=n^{-2/p}$) and true for nonincreasing sequences $(a_n)$ if $p\ge2$ (see second part of the post below).
Rearranging terms, one sees that the double series converges if and only if the simple series $\sum\limits_{n=1}^{+\infty}na_n^p$ does. But this does not need to be the case.
To be specific, choose a positive real number $r$ and let $a_n=0$ for every $n$ not the $p$th power of an integer (see notes) and $a_{i^p}=i^{-(1+r)}$ for every positive $i$. Then $\sum\limits_{n=1}^{+\infty}a_n$ converges because $\sum\limits_{i=1}^{+\infty}i^{-(1+r)}$ does but $na_{n}^p=i^{-pr}$ for $n=i^p$ hence $\sum\limits_{n=1}^{+\infty}na_n^p$ diverges for small enough $r$.
Notes:
(1) If $p$ is not an integer, read $\lfloor i^p\rfloor$ instead of $i^p$.
(2) If the fact that some $a_n$ are zero is a problem, replace these by positive values which do not change the convergence/divergence of the series considered, for example add $2^{-n}$ to every $a_n$.
To deal with the specific case when $(a_n)$ is nonincreasing, assume without loss of generality that $a_n\le1$ for every $n$ and introduce the integer valued sequence $(t_i)$ defined by $ a_n\ge2^{-i} \iff n\le t_i. $ In other words, $ t_i=\sup\{n\mid a_n\ge2^{-i}\}. $ Then $\sum\limits_{n=1}^{+\infty}a_n\ge u$ and $\sum\limits_{n=1}^{+\infty}na_n^p\le v$ with $ u=\sum\limits_{i=0}^{+\infty}2^{-i}(t_i-t_{i-1}),\quad v=\sum\limits_{i=0}^{+\infty}2^{-ip-1}(t_{i+1}^2-t_i^2). $ Now, $u$ is finite if and only if $\sum\limits_{i=0}^{+\infty}2^{-i}t_i$ converges and $v$ is finite if and only if $\sum\limits_{i=0}^{+\infty}2^{-ip}t_i^2$ does. For every $p\ge2$, one sees that $2^{-ip}t_i^2\le(2^{-i}t_i)^2$, and $\ell^1\subset\ell^2$, hence $u$ finite implies $v$ finite.