$\textrm{let us assume $sin^{-1}(\frac{3}{5})=\theta$}$
$\implies$ $\sin\theta=\frac{3}{5}$
$\because \sin\theta=\frac{perpendicular}{base}$
$\textrm{by using pythagoras theorem we can find the base of right angle triangle }$
$\implies$ $H^2=B^2+P^2$
$\textrm{base =4}$
$\therefore$ $\tan\theta=\frac{perpendicular}{base}$
$\implies$ $\tan\theta=\frac{3}{4}$
$\implies$ $\theta=\tan^{-1}\frac{3}{4}$
$\therefore$ $tan^{-1}\frac{3}{4}+tan^{-1}\frac{1}{7}$
$\because \tan^{-1}(A+B)=tan^{-1}(\frac{A+B}{1-AB})$
$\implies$ $\tan^{-1}(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}*\frac{1}{7}})$
$\implies$ $tan^{-1}(1)$
$\because tan^{-1}(1)=\frac{\pi}{4}$
$\textrm{option A is correct.}$