The areal velocity $ \large\frac{dA}{dt}$ =constant
$\large\frac{dA}{dt}=\frac{L}{2 m}$
$L=mvr \sin \theta$ (angular momentum)
$\large\frac{dA}{dt}=\frac{mvr \sin \theta}{2m}$
$\qquad= v r \sin \theta$
$\therefore \large\frac{dA}{dt} $$\; \alpha\; m^0$
Hence b is the correct answer.