Answer: $\textbf D$
Let: $\bf {\alpha} = \sin ^{-1}\frac{1}{\sqrt 5}$, and $\bf {\beta} = \cos^{-1} \frac{3}{\sqrt {10}}$
$\therefore \sin \alpha = \frac{1}{\sqrt 5} = \frac{\bf P}{\bf H}, \; \cos \beta = \frac{3}{\sqrt{10}}= \frac{\bf B} {\bf H}$
By using Pythagoras Theorem:
$(\because \mathrm {P^2 + B^2 = H^2 \Rightarrow B = \sqrt {H^2 - P^2}} = \sqrt {5-1} = 2)$
$\Rightarrow \cos \alpha = \frac{2}{\sqrt 5}$
Similarly,
$\Rightarrow \cos \beta = \frac{3}{\sqrt {10}}$
$\Rightarrow \sin \beta = \frac{1}{\sqrt {10}}$
Now, We know that:
$\mathrm {\sin (A+B) = \sin A \cos B + \cos A \sin B}$
$\begin {align}\Rightarrow \sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta\\ &= \frac{1}{\sqrt 5}.\frac{3}{\sqrt {10}} + \frac{2}{\sqrt {5}}. \frac{1}{\sqrt{10}} \\ &= \frac{3}{\sqrt{50}} + \frac{2}{\sqrt{50}}\\&=\frac{5}{\sqrt{50}}\\&= \frac{5}{5\sqrt 2} \\&=\frac{1}{\sqrt 2}\end {align}$
$\begin {align} \Rightarrow \alpha + \beta &= \sin ^{-1} \frac{1}{\sqrt 2} \\&= \sin ^{-1} \Bigg({\sin \frac{\pi}{4}}\Bigg) \\&= \frac{\pi}{4}\end {align}$
Answer $\therefore \textbf D $ is the right option.
