Answer: Option B
:
B
Let us take the origin at P, X-axis along the horizontal and Y-axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time 0 to t,

$v_x=v_{0}cos{45}^{\circ }=\frac{v_0}{\sqrt{2}}$
and $x=v_{x}t=\frac{v_0}{\sqrt{2}}.\frac{v_0}{g}=\frac{v_0^2}{\sqrt{2}g}$
For vertical motion,
$v_{\gamma }=v_{0}sin{45}^{\circ }-gt=\frac{v_0}{\sqrt{2}}-v_0=\frac{1-\sqrt{2}}{\sqrt{2}}{v}_0$
and $y=\left(v_{0}sin{45}^{\circ }\right)t-\frac{1}{2}g{t}^2$
$y=\frac{v_0^2}{\sqrt{2}g}-\frac{v_0^2}{2g}=\frac{v_0^2}{2g}(\sqrt{2}-1)$
The angular momentum of the particle at time t about the origin is
$L=\vec{r}\times \vec{p}=m\vec{r}\times \vec{v}$
=$m(\overrightarrow{ix}+\overrightarrow{jy})\times (\overrightarrow{i{v}_x}+\overrightarrow{j{v}_y})$
=$m(\vec{k}x{v}_y-\overrightarrow{k_y}{v}_x)$
$=m\vec{k}\left[\left(\frac{v_0^2}{\sqrt{2}g}\right)\frac{v_0}{\sqrt{2}}\right(1-\sqrt{2})-\frac{v_0^2}{2g}(\sqrt{2}-1\left)\frac{v_0}{\sqrt{2}}\right]$
$=-\vec{k}\frac{m{v}_0^3}{2\sqrt{2}g}$
Thus, the angular momentum of the particle is $\frac{m{v}_0^3}{2\sqrt{2}g}$ in the negative Z-direction, i.e., perpendicular to the plane of motion, going into the plane.