If $\large\frac{2^{m+n}}{2^{n-m}}=$$16$ and $a=2^{\large\frac{1}{10}},$ then $\large\frac{a^{2m+n-p}}{(a^{m-2n+2p})^{-1}}$

Question

$( A ) 9 \\ ( B ) 2\\ ( C ) \large\frac{1}{4} \\ ( D ) \large\frac{1}{8}$