Answer: Option B
:
B

As the ballhas to hit the inclined plane normally, so in that position the x-component of velocity will be zero and velocity will have y-component only.
The ball will hit the incline normally if its parallel component of velocity reduces to zero during the time of flight.
By analyzing this motion along incline, i.e., x-direction $v_x=u_x+a_{x}t$
Here $v_x=0,u_x=v_{0}cos\theta ,a_x=-gsin\alpha$
$0=v_{o}cos\theta -\left(gsin\alpha \right)T\Rightarrow T=\frac{v_{o}cos\theta }{gsin\alpha }$ ........(i)
Also the displacement of the particle in y-direction will be zero. Using
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow 0=v_{o}sin\theta .T-\frac{1}{2}gcos\alpha .T^2$
This gives T = $\frac{2v_{o}sin\theta }{gcos\alpha }$ ...............(ii)
From (i) and (ii), we have $\frac{v_{o}cos\theta }{gsin\alpha }=\frac{2v_{o}sin\theta }{gcos\alpha }\Rightarrow \frac{cos\theta }{sin\alpha }=\frac{2sin\theta }{cos\alpha }\Rightarrow 2tan\theta tan\alpha =1\Rightarrow tan\theta =\left[\frac{1}{2}cos\alpha \right]$
$\Rightarrow =ta{n}^{-1}\left(\frac{1}{2}cot\alpha \right)$ which is the required angle of projection.