Answer: Option A
:
A
Let's analyze the motion

The path that the falling ball takes is same. The ball starts from A and hits the ground at B then rebounds on the same path AB to reach A and falls again continuing the motion (A - B - A - B - - - -)
Now in the motion from A to B,
u = 0
s = -h ; a = -g
$s=ut+\frac{1}{2}a{t}^2$
$-h=\frac{-1}{2}g{t}^2$
Time taken to fall t =$\sqrt{\frac{2h}{g}}$
Time taken to go back is same t =$\sqrt{\frac{2h}{g}}$
Total time = 2$\sqrt{\frac{2h}{g}}$ , which is the "time period for this motion”
for (A - B - A)
since the collision was elastic the ball comes back to same point and falls again.
It follows the same path and takes the same time.
Is the motion repeating itself?
Well what do we mean by "motion repeats itself”
(1) Take a reference point (any point)
(2) Note time period of motion. Let it be 'T'
(3) Note particles position and velocity from the reference point at any time t, (t $<$ T)
(4) Again note particles position and velocity from reference point at time (T + t)
(5) The result for (3) $\&$ (4) should be same for a motion to be periodic. Also, the time period of every cycle should be same
The above motion is periodic
Let's check whether it's oscillatory
Oscillatory definition:
Oscillatory Motion:- If the body moves to and fro under the influence of a restoring force then its called oscillatory motion There is no point for stable equilibrium. So no point of restoring force