When a ball hits a wall (ground), some quantity of its momentum must be transferred to the wall at rest.
That is correct.
If else the ball would bounce back with the exact same speed as it did when it collided with the wall. This would result in the ball traveling up to the exact height from which it was dropped to the ground, which is not seen.
That is not necessarily correct. Theoretically the ball could bounce back with exactly the same speed as it did when it collided with the wall or bounce back to the same height after impacting the ground. It would happen if the collision were perfectly elastic. The reason it is not seen is because no collisions are perfectly elastic. All collisions of macroscopic bodies are inelastic, some more inelastic than others. Some kinetic energy is always lost (converted to heat, sound, etc.).
Does this mean that the ground receives momentum but it gets converted into sound energy by the vibration of ground which had collided with the ball?
No.
First of all, it isn’t momentum that is converted into sound, heat or other energy forms. It is the kinetic energy lost in an inelastic collision. Momentum is always conserved.
I think you are assuming the velocity of the ground or the wall are zero after the collision. They are not. It’s just that the mass of the ground or wall, which is essentially that of the earth is so much greater than the ball that their velocities after the collision are too small to be observed.
Whose net result being that: the ball does not deform and the ground vibrates; the conservation momentum principle holds as momentum is transferred to the ground!
Conservation of momentum holds regardless of whether or not the ball deforms or the ground vibrates. Regarding deformation of the ball (or anything else) there basically two types, elastic and plastic. See answer to your question about coefficient of restitution below.
But, can someone explain the coefficient of restitution!
Basically, the coefficient of restitution is the ratio of the total initial kinetic energy of two colliding bodies after a collision to the total kinetic energy of the two colliding bodies before a collision. It is a number between zero for a completely inelastic collision (for example, something hits a wall and sticks to it) to 1 for a perfectly elastic collision. As I indicated above, all collisions at the macroscopic level are inelastic so the coefficient of restitution is always less than 1..
Taking the example of the ball hitting a wall and bouncing back. We know the kinetic energy of the wall is zero before the collision. Technically it is not zero after the collision because of conservation of momentum but as a practical matter it is considered zero when considering the coefficient of restitution of the ball.
The following responds to these follow up questions:
I was actually relating the conservation of momentum principle and the equations of kinematics! Is it better not to do that?
Kinematics is basically the analysis of motion in terms of displacement, time, velocity, and acceleration without regard to the forces involved that cause motion. It's OK to relate it to conservation of momentum since certainly velocity plays the primary role (in addition to mass) relating to conservation of momentum.
I'm also assuming the momenta of the wall before or after the collision are both zero... Where did I go wrong?
As I already indicated, assuming the momentum of the wall after the collision is zero is incorrect. Consider the following example.
A pitched baseball impacts a wall. It has a velocity of 40 m/s prior to impact (90 mph). The mass of the baseball is about 0.145 kg. Consider the wall to be an extension of the earth. The mass of the earth about 6 x $10^{24}$ kg. Let the collision of the ball be perfectly elastic, so its rebound velocity is the same. Since the ball reverses itself its change in momentum is -(0.145)(40)= -5.8 kg-m/s.
For conservation of momentum the change is momentum of the earth must be equal and opposite of the ball, or +5.8 kg-m/s. So the velocity of the earth following the collision is +5.8 kg-m/s divided by the mass of the earth, or V= 9.75 x $10^{-25}$ m/s.
Hope this helps.