A solid sphere, a hollow sphere and a disc, all having the same...

Since the friction is not sufficient to allow pure rolling, all the objects will slip as they move down the incline. The kinetic friction force acting on each object is $f_k = \mu M g \cos\theta$.

The linear acceleration of the center of mass for all objects is given by:
$a = g \sin\theta - \mu g \cos\theta$

Since $a$ is independent of the shape and moment of inertia, all three objects will take the same time $t$ to reach the bottom and will have the same translational velocity $v = at$. Thus, their translational kinetic energies $K_{\text{trans}} = \dfrac{1}{2} M v^2$ will be identical.

The angular acceleration of each object is:
$\alpha = \dfrac{\tau}{I} = \dfrac{f_k R}{I}$

The angular velocity at the bottom of the incline is:
$\omega = \alpha t = \dfrac{f_k R t}{I}$

The rotational kinetic energy at the bottom is:
$K_{\text{rot}} = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} I \left( \dfrac{f_k R t}{I} \right)^2 = \dfrac{(f_k R t)^2}{2 I}$

The total kinetic energy is:
$K = K_{\text{trans}} + K_{\text{rot}} = \dfrac{1}{2} M v^2 + \dfrac{(f_k R t)^2}{2 I}$

Since $M$, $v$, $f_k$, $R$, and $t$ are the same for all three objects, the total kinetic energy $K$ is inversely related to the moment of inertia $I$. To find the smallest kinetic energy, we need the object with the largest moment of inertia.

The moments of inertia for the given objects are:
Solid sphere: $I = \dfrac{2}{5} M R^2 = 0.4 M R^2$
Disc: $I = \dfrac{1}{2} M R^2 = 0.5 M R^2$
Hollow sphere: $I = \dfrac{2}{3} M R^2 \approx 0.67 M R^2$

The hollow sphere has the largest moment of inertia, which means it will acquire the smallest rotational kinetic energy. Therefore, the hollow sphere will achieve the smallest total kinetic energy at the bottom of the incline.

Answer: the hollow sphere