Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X-$ axis and $P\left(5,y_1\right)$ be point on the hyperbola. Then $\mathrm{SP}=$

Since, point $P\left(5,y_1\right)$ lies on the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, therefore

$\frac{5^2}{16}-\frac{y_1^2}{9}=1$

$\Rightarrow \frac{y_1^2}{9}=\frac{25}{16}-1$

$\Rightarrow y_1^2=\frac{81}{16}$

$\Rightarrow y_1=\pm \frac{9}{4}$

So, $P\equiv \left(5,\pm \frac{9}{4}\right)$.

Now, eccentricity of hyperbola is

$e=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$

Therefore, $S\equiv \left(5,0\right)$

Hence,

$SP=\sqrt{{\left(5-5\right)}^2+{\left(0-\frac{9}{4}\right)}^2}=\frac{9}{4}$