Let $g$ be the acceleration due to gravity. The driver undergoes uniform circular motion, so the net force on the driver provides the centripetal force $F_c = mv^2/R$.

At the top of the hill, the gravitational force $mg$ acts downwards and the normal force $N_{top}$ acts upwards. The net force is directed downwards, towards the center of the circle. Applying Newton's second law:

$$mg - N_{top} = \frac{mv^2}{R}$$

Given that $N_{top} = 0$, the equation simplifies to:

$$mg = \frac{mv^2}{R} \implies v^2 = gR$$

At the bottom of the valley, the gravitational force $mg$ acts downwards and the normal force $N_{bottom}$ acts upwards. The net force is directed upwards, towards the center of the circle. Applying Newton's second law:

$$N_{bottom} - mg = \frac{mv^2}{R}$$

Substitute the expression for $v^2$ found from the condition at the top of the hill:

$$N_{bottom} = mg + \frac{m(gR)}{R} = mg + mg$$ $$N_{bottom} = 2mg$$