1. Use standard results for a planet in circular orbit of radius $a$ around the Sun

For a planet of mass $m$ orbiting a Sun of mass $M$:

• Gravitational potential energy: $U=-\frac{GMm}{a}$
• From centripetal force, $\frac{mv^2}{a}=\frac{GMm}{a^2}\Rightarrow v^2=\frac{GM}{a}$
• Hence kinetic energy: $K=\frac12 mv^2=\frac12 m\cdot \frac{GM}{a}=\frac{GMm}{2a}$
• Total mechanical energy: $E=K+U=\frac{GMm}{2a}-\frac{GMm}{a}=-\frac{GMm}{2a}$

2. Escape energy at the surface of the planet for unit mass object

Escape energy for unit mass from a planet of mass $m$ and radius $r$ is the minimum energy needed to make total energy zero:

$E_{\text{escape}}=\frac{Gm}{r}$

3. Now match List I with List II

• A. Kinetic energy of planet $K=\frac{GMm}{2a}$ So, A $\to$ II

• B. Gravitational potential energy of sun-planet system $U=-\frac{GMm}{a}$ So, B $\to$ I

• C. Total mechanical energy of planet $E=-\frac{GMm}{2a}$ So, C $\to$ IV

• D. Escape energy at the surface of planet for unit mass object $E_{\text{escape}}=\frac{Gm}{r}$ So, D $\to$ III

4. Final matching

$A\to II,\quad B\to I,\quad C\to IV,\quad D\to III$

This corresponds to Option A.