1. Electronic configurations

• Be $(Z=4)$: $1s^2 2s^2$
• B $(Z=5)$: $1s^2 2s^2 2p^1$

The first ionization enthalpy of B is smaller than that of Be because the electron removed from B is a $2p$ electron, while from Be it is a $2s$ electron.

2. Examine Statement (I):

It is easier to remove $2p$ electron than $2s$ electron.

This is true.

Reason: A $2p$ electron is higher in energy and less penetrating than a $2s$ electron, so it is held less strongly by the nucleus.

3. Examine Statement (II):

$2p$ electron of B is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of Be.

This is true in the intended periodic-trend sense.

Reason: The $2p$ electron is less penetrating and spends less time close to the nucleus than a $2s$ electron. Hence it experiences greater shielding and lower effective nuclear charge than a $2s$ electron of comparable shell.

4. Examine Statement (III):

$2s$ electron has more penetration power than $2p$ electron.

This is true.

Known order of penetration: $s > p > d > f$ So $2s$ penetrates closer to the nucleus than $2p$.

5. Examine Statement (IV):

Atomic radius of B is more than Be.

This is false.

Across a period, atomic radius generally decreases from left to right due to increasing effective nuclear charge. Thus, $r(\text{B}) < r(\text{Be})$ So this statement is incorrect.

6. Correct combination

True statements are:

• (I) True
• (II) True
• (III) True
• (IV) False

Therefore, the correct option is: $\boxed{\text{D: (I), (II) and (III)}}$