$y=ax-bx^2$
for maximum height
$\large\frac{dy}{dx}$$=0$
$a-2bx=0$
$x=\large\frac{a}{2b}$
Therefore $y_{\large max}=a \bigg(\large\frac{a}{2b}\bigg)$$-b\bigg(\large\frac{a}{2b}\bigg)^2=\frac{a^2}{4b}$
angle of projection $\tan {\theta}=> at \;x=0$
$\bigg(\large\frac{dy}{dx}\bigg)_{x=0}=\large\frac{d}{dx}$$[ax-bx^2]\bigg|_{x=0}$
$\quad\qquad\qquad=a-2bx \bigg|_{x=0}$
$\qquad\qquad\quad=a$
Therfore $\tan \theta =a$
$\qquad \theta=\tan ^{-1} (a)$
Hence b is the correct answer.