Given curve is
$x=a(\theta+\sin \theta), y=a(1-\cos \theta)$
$\therefore \quad \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}=\tan \frac{\theta}{2}$
$\therefore$ Slope of tangent and normal drawn to the given curve at $P\left(\theta=\frac{\pi}{2}\right)$ is $m_T=1$ and $m_N=-1$ respectively.
Now, equation of tangent to the given curve at point $P\left(a\left(\frac{\pi}{2}+1\right), a\right)$ is $y-a=1[x-a(\pi / 2+1)]$
$\therefore$ Point $A$ is $\left(\frac{a \pi}{2}, 0\right)$
and equation of normal to the given curve at point $P$ is $y-a=-1\left[x-a\left(\frac{\pi}{2}+1\right)\right]$
So, point $B$ is $\left(a\left(\frac{\pi}{2}+2\right), 0\right)$
Therefore area of $\triangle P A B=\frac{1}{2}\left|\begin{array}{ccc}a\left(\frac{\pi}{2}+1\right) & a & 1 \\ \frac{a \pi}{2} & 0 & 1 \\ a\left(\frac{\pi}{2}+2\right) & 0 & 1\end{array}\right|$
$=\frac{1}{2}\left|a\left[\frac{a \pi}{2}-\frac{a \pi}{2}-2 a\right]\right|=a^2$