We have,

$x^2+y^2-8x-10y+5=0$

$\Rightarrow x^2+y^2-8x-10y+16+25=16+25-5$

$\Rightarrow {\left(x-4\right)}^2+{\left(y-5\right)}^2=6^2$

Centre $C\equiv \left(4,5\right)$

Radius$=CA=CB=6$ units.

Now,

$PA=PB=\sqrt{S_1}=\sqrt{4+9+16+30+5}=8$

In $∆CAP$,

$\frac{CA}{PC}=\tan \left(\frac{\theta }{2}\right)$

$\Rightarrow \tan \left(\frac{\theta }{2}\right)=\frac{6}{8}=\frac{3}{4}$

Now,

$\tan \theta =\frac{2\tan \left({\displaystyle \frac{\theta }{2}}\right)}{1-{\tan }^2\left({\displaystyle \frac{\theta }{2}}\right)}=\frac{{\displaystyle \frac{6}{4}}}{1-{\displaystyle \frac{9}{16}}}$

$\Rightarrow \tan \theta =\left(\frac{{\displaystyle \frac{6}{4}}}{{\displaystyle \frac{7}{16}}}\right)=\frac{24}{7}$