You can find the question in the attachments, that's because I don't know how to type the equation in a perfect way by using the keyboard.
Start by drawing a graph. The outer integral, with respect to x goes from 0 to p so draw vertical lines at x= 0 (the y-axis) and at x= p. The inner integral, with respect to y, goes from $\displaystyle \sqrt{px}$ up to p so draw the graph of $\displaystyle y= \sqrt{px}$, part of a parabola with horizontal axis and the horizontal line y= p. Note that the axes, x= p, and y= p form a square and, since $\displaystyle y(p)= \sqrt{p^2}= p$ (for p> 0), the parabola is kind of curved "diagonal" of that square. The inner integral is from $\displaystyle \sqrt{px}$ to p so the area of integration is the region above that diagonal. To cover that area, y must go from 0 to p and, for each y, x must go from 0 to the parabola $\displaystyle y= \sqrt{px}$, $\displaystyle y^2= px$, $\displaystyle x= y^2/p$.
The integral is $\displaystyle \int_0^p\int_0^{y^2/p} \frac{y^2}{\sqrt{y^4- p^2x^2}}dxdy$
The first integral is $\displaystyle y^2\int_0^{y^2/p} \frac{dx}{\sqrt{y^4- p^2x^2}}$$\displaystyle = y\int_0^{y^2/p}\frac{1}{\sqrt{1- \frac{p^2x^2}{y^4}}}dx$
That should make you think of a "trig" substitution: Let $\displaystyle \frac{px}{y^2}= sin(\theta)$. Then $\displaystyle \frac{p}{y^2}dx= cos(\theta)d\theta$, $\displaystyle \frac{1}{\sqrt{1- \frac{p^2x^2}{y^4}}}= \frac{1}{\sqrt{1- sin^2(\theta)}}= \frac{1}{cos(\theta)}$, and, when $\displaystyle x= \frac{y^2}{p}$, $\displaystyle sin(\theta)= 1$ so that the integral becomes $\displaystyle \int_0^{\pi/2} \frac{cos(\theta) d\theta}{cos(\theta)}$$\displaystyle = \int_0^{\pi/2} d\theta$ which is independent of y!