D

Step 1: Consider the charge +q thrown towards the charged disc. The disc has a uniform charge distribution, which creates an electric field that varies depending on its distance from the disc.
Step 2: Analyzing the kinetic energy of the charge +q, given as K.E. = \frac{1}{2} mv^2, it can approach the disc with sufficient energy to either reach the disc, be repelled, or return back.
Step 3: The electric field (E) created by the disc will exert forces on the charge +q depending on its proximity and speed.
Step 4: If the kinetic energy is more than the work done against the electric field, it will hit the disc. If it's less than the work done, it will return back. Thus, any of the three options is possible depending on the magnitude of E and the initial kinetic energy.
Therefore, the correct option is D.