jmc

Notice that if we add $a^4$ to both sides of the first equation, we get $a^{7}xy-a^{6}y-a^{5}x+a^4=a^4{b}^4$. Factoring the left side gives $(a^{3}x-a^2)(a^{4}y-a^2)=a^4{b}^4$. So, $(m,n,p)=(3,2,4)$, which means $mnp=3\cdot 2\cdot 4=\boxed{24}$.