Given hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

$P\left(3,3\right)$ lies on hyperbola then $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{9} ...\left(1\right)$

Normal at $\left(3, 3\right)$ is $\frac{a^{2}x}{3}+\frac{b^{2}y}{3}=a^2+b^2$

Normal passes through $\left(9, 0\right)$

$3a^2=a^2+b^2 \Rightarrow 2a^2=b^2$

Then using $\left(1\right)$, $\frac{1}{a^2}-\frac{1}{2a^2}=\frac{1}{9}$

$2a^2=9\Rightarrow a^2=\frac{9}{2}$ and $b^2=9$

$e^2=1+\frac{b^2}{a^2}=1+2=3$.