We discuss the options as follows:

$$\bullet$$ For option (B) : From law of conservation of momentum, we have

mv = MV ...... (1)

where v is the velocity of point mass m and V is velocity of block of mass M.

Now, by using law of conservation of energy, after the point mass is released, we have the following:

Loss in P.E. of mass m = Gain in K.E. of both mass M and the mass m.

$$ \Rightarrow mgh = {1 \over 2}m{v^2} + {1 \over 2}M{V^2}$$

Substituting $$V = {{mu} \over M}$$ [from Eq. (1)], we get

$$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{\left( {{{mv} \over M}} \right)^2}$$

$$mgh = {1 \over 2}m{v^2} + {1 \over 2}M{{{m^2}{v^2}} \over {{M^2}}}$$

$$mgR = {1 \over 2}m{v^2} + {1 \over 2}{{{m^2}{v^2}} \over M}$$

$$mgR = {1 \over 2}m{v^2}\left( {1 + {m \over M}} \right)$$

$$gR = {1 \over 2}{v^2}\left( {1 + {m \over M}} \right)$$

Rearranging this equation, we get

$$2gR = {v^2}\left( {1 + {m \over M}} \right)$$

$${v^2} = {{2gR} \over {\left( {1 + {m \over M}} \right)}}$$

$$ \Rightarrow v = \sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $$

Hence, option (B) is correct.

$$\bullet$$ For option (C) : Now, from Eq. (1), we have

$$V = {m \over M}v = {m \over M}\sqrt {{{2gR} \over {\left( {1 + {m \over M}} \right)}}} $$

$$V = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {1 + {m \over M}} \right)}}} = \sqrt {{{{m^2}2gR} \over {{M^2}\left( {{{M + m} \over M}} \right)}}} $$

$$V = \sqrt {{{{m^2}2gR} \over {M(m + M)}}} \Rightarrow V = m\sqrt {{{2gR} \over {M(m + M)}}} $$

Since, there is no external force acting on the system, the centre of mass does not change.

Now, if the change in position of mass m is $$\Delta$$x and the change in position of mass M is $$\Delta$$X, then

m$$\Delta$$x + M$$\Delta$$x = 0 (since centre of mass will not change)

Now, we know that the change in position of point mass m is

$$\Delta$$x = R $$-$$ x

and the change in position of block of mass M is

$$\Delta$$X = $$-$$x

Therefore,

m(R $$-$$ x) $$-$$ Mx = 0

m(R $$-$$ x) = Mx

mR $$-$$ mx = Mx

mR = Mx + mx

mR = (M + m)x

$$\Rightarrow$$ x = $${{mR} \over {M + m}}$$

The change in position of block of mass M is

$$\Delta$$x = $$-$$x = $${{ - mR} \over {M + m}}$$

Thus, x-component of the displacement of the centre of mass of the block is

$$M = {{ - mR} \over {M + m}}$$

Hence, option (C) is also correct.