Let me prove it for $p=3$ (only because the notations look more friendly.) The maximum of left hand side is realized when the arrays $(a_i),(b_{\sigma(i)}),(c_{\tau(i)})$ are equally sorted, so let us assume that it is the case.
I claim that not only the sum, but the $s$-th largest summand of RHS (clarification: first largest means maximal) is not less than the $s$-th largest summand in LHS, for any $s=1,2,\dots,n$. Denote by $\alpha, \beta, \gamma$ the $s$-th largest elements of the arrays $a, b, c$ respectively. Choose minimal indices $i, j, k$ respectively for which $a_i\geqslant \alpha$, $b_j\geqslant \beta$, $c_k\geqslant \gamma$. Without loss of generality $k=\max(i, j, k) $. Then $m_r\geqslant \alpha \beta \gamma$ for all $r$ such that $c_r\geqslant \gamma$ (because all such $r$ are not less than $\max(i, j)$), and we get at least $s$ such values of $r$, thus $s$-th largest $m$ is not less than $\alpha \beta\gamma$, as desired.