A ball is released from rest from point P of a smooth...

At angular position $\alpha$ as shown in the figure, applying conservation of mechanical energy

$mgRsin\alpha = \dfrac{1}{2}m{v}^{2}$
$⇒ \dfrac{m{v}^{2}}{R} = 2mgsin\alpha$

In the radial direction,

$N - mgsin\alpha = \dfrac{m{v}^{2}}{R}$
$⇒ N = 3mgsin\alpha$

Now the ratio of centripetal force and normal reaction is,

$A = \dfrac{\dfrac{m{v}^{2}}{R}}{N} = \dfrac{2mgsin\alpha}{3mgsin\alpha} = \dfrac{2}{3}$.  Therefore, $A$ is constant.