$\frac{x^2}{25}-\frac{y^2}{9}=1$ is the given hyperbola.

Any point on the hyperbola is given by $\left(5\sec \theta ,3\tan \theta \right)$

Given point on the parabola is $P\left(\theta \right)=\left(x_1,\frac{3\sqrt{5}}{2}\right),0<\theta <\frac{\pi }{2}$

i.e. $\tan \theta =\frac{\sqrt{5}}{2}, \sec \theta =\frac{3}{2}$

i.e. $x_1=\frac{15}{2}$

So, $2x_1+9{\sin }^2\theta =15+9{\left(\frac{\sqrt{5}}{3}\right)}^2=20$