(d)

Since, $\angle D A B, \angle A B C$ and $\angle B C D$ are in $AP \therefore$ Let $\angle D A B=\theta-\alpha, \angle A B C=\theta$ and $\angle B C D=\theta+\alpha$

$\therefore$ Median of $\angle D A B, \angle A B C$ and $\angle B C D=\theta$

From point $E$ all the vertices are at equal distance.

$\therefore A B C D$ is cyclic.

and $\angle A D C=2 \pi-(\theta-\alpha+\theta+\theta+\alpha)$

$\quad=2 \pi-3 \theta$

and $\angle A D C+\angle A B C=\pi$

$\Rightarrow 2 \pi-3 \theta+\theta=\pi$

$\therefore \quad \theta=\frac{\pi}{2}$