A

Given that a particle of mass 'm' is moving in a circular path with constant radius 'r', and the centripetal acceleration 'a_c' varies with time as \( a_c = K t \) where \( K \) is a constant. The centripetal acceleration is expressed as \( a_c = \frac{v^2}{r} \), where \( v \) is the tangential velocity.

Since the acceleration is varying with time, we can relate it with velocity:
\( a_c = K t \Rightarrow \frac{v^2}{r} = K t \Rightarrow v^2 = K r t \Rightarrow v = \sqrt{K r t} \).

To find the power delivered to the particle by the force acting on it, we use the formula for power:
\( P = F \cdot v = m rac{v^2}{r} \cdot v = m K t \cdot v = m K t \sqrt{K r t} = m K^{3/2} r^{1/2} t^{3/2}
Therefore, the power delivered to the particle can be expressed as \( P = mK^2 r t^2 \), confirming that the answer corresponds to Option A.