Answer
$-\dfrac{3}{32m^{8}p^{4}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\dfrac{(2m^2p^3)^2(4m^2p)^{-2}}{(-3mp^4)^{-1}(2m^3p^4)^3}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(2^2m^{2(2)}p^{3(2)})(4^{-2}m^{2(-2)}p^{-2})}{(-3)^{-1}(m^{-1}p^{4(-1)})(2^3m^{3(3)}p^{4(3)})}
\\\\=
\dfrac{(4m^{4}p^{6})(4^{-2}m^{-4}p^{-2})}{(-3)^{-1}(m^{-1}p^{-4})(8m^{9}p^{12})}
\\\\=
\dfrac{4(-3)^{1}m^{4+(-4)}p^{6+(-2)}}{4^{2}(m^{-1+9}p^{-4+12})(8)}
\\\\=
\dfrac{4(-3)m^{0}p^{6-2}}{16(m^{8}p^{8})(8)}
\\\\=
\dfrac{-12m^{0}p^{4}}{128m^{8}p^{8}}
\\\\=
\dfrac{\cancel{4}(-3)m^{0-8}p^{4-8}}{\cancel{4}(32)}
\\\\=
\dfrac{-3m^{-8}p^{-4}}{32}
\\\\=
-\dfrac{3}{32m^{8}p^{4}}
.\end{array}