Answer
$\dfrac{1}{p^{2}(m+n)}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
\dfrac{[p^2(m+n)^3]^{-2}}{p^{-2}(m+n)^{-5}}
.$
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{p^{2(-2)}(m+n)^{3(-2)}}{p^{-2}(m+n)^{-5}}
\\\\=
\dfrac{p^{-4}(m+n)^{-6}}{p^{-2}(m+n)^{-5}}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
p^{-4-(-2)}(m+n)^{-6-(-5)}
\\\\=
p^{-4+2}(m+n)^{-6+5}
\\\\=
p^{-2}(m+n)^{-1}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{p^{2}(m+n)^{1}}
\\\\=
\dfrac{1}{p^{2}(m+n)}
.\end{array}
