Initial Conditions:

Consider a solid disc of mass $ M $ and radius $ R $ rolling without slipping on a horizontal surface. The center of mass of the disc has an initial speed $ v $.

Kinetic Energy of the Rolling Disc:

For an object that rolls without slipping, its total kinetic energy $ K $ is the sum of its translational kinetic energy and its rotational kinetic energy.

Translational Kinetic Energy:

$ K_{\text{trans}} = \frac{1}{2} M v^2. $

Rotational Kinetic Energy:

The moment of inertia $ I $ of a solid disc about its center is:

$ I = \frac{1}{2} M R^2. $

Since the disc rolls without slipping, the angular velocity $\omega$ is related to the linear speed by:

$ v = \omega R \implies \omega = \frac{v}{R}. $

Therefore, the rotational kinetic energy is:

$ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v^2}{R^2}\right) = \frac{1}{4} M v^2. $

Total Initial Kinetic Energy:

Summing these gives:

$ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2. $

Energy Conservation as the Disc Climbs the Incline:

As the disc moves up the incline, there are no dissipative forces (assuming a smooth incline and no slipping), so energy is conserved. The total kinetic energy at the bottom will be converted entirely into gravitational potential energy at the maximum height $ h $:

$ K_{\text{total}} = M g h. $

Substitute $ K_{\text{total}} = \frac{3}{4} M v^2 $:

$ \frac{3}{4} M v^2 = M g h. $

Canceling $ M $ from both sides:

$ \frac{3}{4} v^2 = g h. $

Thus,

$ h = \frac{3}{4} \frac{v^2}{g}. $

Final Answer:

$ \boxed{\frac{3}{4} \frac{v^2}{g}} $

This corresponds to Option A.