\(\left(\dfrac{9a^{-4}}{25}\right)^p=\dfrac3{5a^2}\)
\(\left(\dfrac9{25a^4}\right)^p=\dfrac3{5a^2}\)
One way to find \(p\) is to notice that the part on the left hand side within the parentheses can be written as a square:
\(\dfrac9{25a^4}=\dfrac{3^2}{5^2(a^2)^2}=\left(\dfrac3{5a^2}\right)^2\)
So you could write
\(\left(\left(\dfrac3{5a^2}\right)^2\right)^p=\left(\dfrac3{5a^2}\right)^{2p}=\dfrac3{5a^2}\)
Since these two expressions are equal, and the bases are the same, you know the exponents must be equal, with the exponent on the right hand side being 1. So \(2p=1\), or \(p=\dfrac12\).
