Explanation

(d)
Given curves are
$x = a(\theta + sin\theta),y = a(1 - cos\theta)$
$\therefore\ \ \ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$
$= \frac{a(\sin\theta)}{a(1 + cos\theta)}$
$= \frac{2\sin{\frac{\theta}{2}\cos\frac{\theta}{2}}}{2\cos^{2}\frac{\theta}{2}}$
$= tan\frac{\theta}{2}$
$\Rightarrow \ \ \ \ \left( \frac{dy}{dx} \right)_{\left( \theta = \frac{\pi}{2} \right)} = \tan{\frac{\pi}{4} = 1}$
At $\theta = \frac{\pi}{2},y = a\left( 1 - cos\frac{\pi}{2} \right) = a$
$\therefore$ length of normal $= y\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}$
$= a\sqrt{1 + (1)^{2}} = \sqrt{2}a$