As shown in the figure, a smooth track includes a circular section...

Let the initial compression of the spring be $x$.

According to the law of conservation of mechanical energy, the initial elastic potential energy of the spring is converted into the gravitational potential energy and kinetic energy of the block at point $P$.

$\dfrac{1}{2}kx^2 = mgR + \dfrac{1}{2}mv^2$

At point $P$, the radius is horizontal, so the normal force $N$ provided by the track acts horizontally towards the center of the circular path. This normal force provides the necessary centripetal force.

$N = \dfrac{mv^2}{R}$

Given that the block presses against the track with a force $mg$, the normal force is $N = mg$.

$mg = \dfrac{mv^2}{R} \Rightarrow v^2 = gR$

Substituting $v^2 = gR$ into the energy conservation equation:

$\dfrac{1}{2}kx^2 = mgR + \dfrac{1}{2}m(gR)$

$\dfrac{1}{2}kx^2 = \dfrac{3}{2}mgR$

$x^2 = \dfrac{3mgR}{k}$

$x = \sqrt{\dfrac{3mgR}{k}}$