Answer
Proved in below
Work Step by Step
$\;\;\;\;\int_0^\infty P(r)dr\\
=\int_0^\infty \frac{4}{a^3}r^2e^{-2r/a}dr\\
=\frac{4}{a^3}\int_0^\infty r^2e^{-2r/a}dr$
Using the following formula
$\int_0^\infty x^ne^{-ax}dx=\frac{n!}{a^{n+1}}$
$\;\;\;\;\int_0^\infty P(r)dr\\
=\frac{4}{a^3}\times\frac{2!}{(\frac{2}{a})^{2+1}}\\
=\frac{4}{a^3}\times\frac{2\times a^{3}}{8}\\
=1$
