1. Treat the block + rope as one system

The total mass being accelerated is $M + m.$

Since the surface is frictionless, the only external horizontal force on the system is $P$.

So the acceleration of the whole system is $a = \frac{P}{M+m}.$

2. Find the force exerted by the rope on the block

The block of mass $M$ is pulled only by the tension at the end of the rope attached to it. Let this force be $T$.

For the block alone, by Newton's second law: $T = Ma.$

Substitute $a = \frac{P}{M+m}$: $T = M\left(\frac{P}{M+m}\right) = \frac{PM}{M+m}.$

3. Match with the options

Thus, the force exerted by the rope on the block is $\boxed{\frac{PM}{M+m}}.$

This corresponds to Option D.

4. Check of other options

• A: $\frac{Pm}{M+m}$ gives the force needed to accelerate mass $m$ proportionally, not the block.
• B: $\frac{Pm}{M-m}$ is not physically relevant here.
• C: $P$ would be correct only if the rope were massless; here the rope has mass, so tension is not equal to $P$ throughout.

Therefore, the correct answer is D.