I don't understand the answer of a) ,please explain what does p.d. drop increases mean.?

The circuit consists of two resistors, the variable resistor R and the internal resistance of the battery r, connected in series. Hence, by Ohm's law,
e = I.(R + r) where e is the emf of the battery and I is the current.
i.e. e = I.R + I.r
But the term (I.R) is the potential difference (p.d.) across the variable resistor R, which is connected to the two terminals of the battery. Thus (I.R.) equals to the terminal voltage (V) of the battery.

Therefore, we have, e = V + I.r
or V = e - I.r

It is just a matter of mathematics that when I decreases, the value of the term (I.r) decreases. Since emf e is fixed, this resulted that V increases.

The physical meaning is that when the value of the variable resistor R is large compared with the internal resistance r, this makes the current to decrease (as the total resistance of the circuit is high). Under such situation, R shares a larger portion of the emf of the battery than r does. Because the voltage across R equals to the terminal voltage of the battery V, this indicates that V will increase.