A

Step 1: Understand the system
We have two planets A and B with masses $M_A$ and $M_B$, respectively, separated by a distance $d$. A satellite of mass $M_S$ is projected from the midpoint between the two planets.

Step 2: Determine the gravitational forces
At the midpoint (distance $\frac{d}{2}$ from both planets), the gravitational forces acting on the satellite due to both planets can be calculated using Newton's law of gravitation:
$$F_A = \frac{G M_A M_S}{(\frac{d}{2})^2} = \frac{4 G M_A M_S}{d^2}$$
$$F_B = \frac{G M_B M_S}{(\frac{d}{2})^2} = \frac{4 G M_B M_S}{d^2}$$
Where $G$ is the universal gravitational constant.

Step 3: Net gravitational force at the midpoint
The net force acting on the satellite is directed towards the larger mass. Hence, if $M_A > M_B$:

$$F_{net} = F_A - F_B = \frac{4 G M_A M_S}{d^2} - \frac{4 G M_B M_S}{d^2} = \frac{4 G M_S (M_A - M_B)}{d^2}$$

Step 4: Gravitational potential energy
The potential energy of the satellite when projected from the midpoint to infinity can be written as:

$$U_{initial} = -\frac{G M_A M_S}{\frac{d}{2}} - \frac{G M_B M_S}{\frac{d}{2}} = -\frac{2 G M_A M_S}{d} - \frac{2 G M_B M_S}{d} = -\frac{2 G M_S (M_A + M_B)}{d}$$

Step 5: Energy conservation
To escape the gravitational pull of both planets to infinity, the total mechanical energy (kinetic + potential) must be zero:

$$K + U_{initial} = 0$$
Where the kinetic energy ($K$) is given by: $$K = \frac{1}{2} M_S v^2$$
Setting the energies equal: $$\frac{1}{2} M_S v^2 - \frac{2 G M_S (M_A + M_B)}{d} = 0$$

Step 6: Solve for minimum velocity
Solving for $v$:
$$\frac{1}{2} M_S v^2 = \frac{2 G M_S (M_A + M_B)}{d}$$
Cancelling out $M_S$ from both sides (as long as $M_S \neq 0$):
$$\frac{1}{2} v^2 = \frac{2 G (M_A + M_B)}{d}$$

Thus, we find
$$v^2 = \frac{4 G (M_A + M_B)}{d}$$
Giving us:
$$v = \sqrt{\frac{4 G (M_A + M_B)}{d}}$$

Hence, the minimum velocity required for the satellite to escape to infinity is:
$$v_{min} = \sqrt{\frac{4G (M_A + M_B)}{d}}$$
Therefore, the answer is option A.