A and  B are two points whose position vectors are  $-\sqrt{3}\widehat{i}\text{ and }-\sqrt{3}\widehat{i}+2\widehat{k}$.Let ${\mathrm{x}}^2+3{\mathrm{y}}^2=3$be the equation of an ellipse  in the XY plane.Then the position vector of a point P on the ellipse such that  $\mathrm{\angle }\mathrm{APB}=\mathrm{\pi }/4$ is 

Point P lies on  ${\mathrm{x}}^2+3{\mathrm{y}}^2=3$------(i)

Now from the diagram, according to the given conditions, AP= AB 

$\text{ or }(\mathrm{x}+\sqrt{3}{)}^2+(\mathrm{y}-0{)}^2=4\text{ or }(\mathrm{x}+\sqrt{3}{)}^2+{\mathrm{y}}^2=4----\left(\mathrm{ii}\right)$

Solving (i) and (ii), we get x = 0 and y =$\pm$ 1 

Hence, point P has position vector $\pm \hat{\mathrm{j}}$