Let $\theta$ be the eccentric angle of the point P.
Then the co-ordinate of P are $(\sqrt 6 \cos \theta, \sqrt 2 \sin \theta)$
The center of the ellipse is at the point of origin.
It is given that
$\therefore OP=2$
=> $ \sqrt {6 \cos^2 \theta + 2\sin ^2 \theta}=2$
=> $ 6 \cos ^2 \theta +2 \sin ^2 \theta=4$
=> $ 3 \cos ^2 \theta + \sin ^2 \theta=2$
=> $ 2 \sin ^2 \theta=1$
=> $\sin ^2 \theta=\large\frac{1}{2}$
=> $\sin \theta= \pm \large\frac{1}{\sqrt 2}$
$\theta= \pm \large\frac{\pi}{4}$
Hence A is the correct answer.