To determine the change in kinetic energy (\(\Delta K\)) of the ball in both frames, we need to analyze the motion of the ball in both the Earth frame (frame \(S\)) and the moving frame (frame \(S'\)).

1. Frame \(S\) (Earth's Frame):
   • The ball is dropped from rest, so its initial velocity \(u_0 = 0\).
   • Using the equation of motion, the final velocity when it hits the ground is:
     \(v_f = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity.
   • The change in kinetic energy is given by:
     \(\Delta K_S = \frac{1}{2}m(v_f^2 - u_0^2) = \frac{1}{2}m(2gh) = mgh\).
2. Frame \(S'\) (Moving Frame):
   • The initial velocity of the ball is \(-u\) (downwards).
   • Applying the relative motion concept and accounting for the frame's upward motion, the velocity of the ball relative to the moving frame just before hitting the ground is:
     \(v_f' = v_f - u = \sqrt{2gh} - u\).
   • The change in kinetic energy in frame \(S'\) is:
     \(\Delta K_{S'} = \frac{1}{2}m((v_f')^2 - (-u)^2)\)
     = \(\frac{1}{2}m((\sqrt{2gh} - u)^2 - u^2)\).
   • After simplifying, the change \(\Delta K_{S'}\) comes out to be \(mgh\), which matches \(\Delta K_S\).

Thus, the change in kinetic energy of the ball is the same in both frames \(S\) and \(S'\). Hence, the correct answer is:

\(\Delta K\) is the same in both the frames.