Q.Assertion (A): Boron has a smaller first ionisation enthalpy than beryllium.
Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron hence 2p electron is more shielded by the inner core of electrons than the 2s electrons.

The key idea is that ionisation enthalpy depends on which orbital the outgoing electron leaves, and how well that orbital penetrates toward the nucleus.

Step 1 — check the assertion. Beryllium ($1s^2\,2s^2$) has $\Delta_i H_1 \approx 899$ kJ mol$^{-1}$; boron ($1s^2\,2s^2\,2p^1$) has $\Delta_i H_1 \approx 801$ kJ mol$^{-1}$. Boron's value IS smaller — the assertion is a correct statement.

Step 2 — check the reason. A $2s$ electron penetrates closer to the nucleus than a $2p$ electron, so the $2p$ electron is more shielded by the inner core and held less tightly. This is also a correct statement. …

Assertion (A): Boron ($1s^2\,2s^2\,2p^1$) does have a smaller first ionisation enthalpy than beryllium ($1s^2\,2s^2$). Measured values are $\Delta_i H_1(\text{Be}) \approx 899\ \text{kJ mol}^{-1}$ versus $\Delta_i H_1(\text{B}) \approx 801\ \text{kJ mol}^{-1}$. So the assertion is a correct statement.

Reason (R): A $2s$ electron penetrates closer to the nucleus than a $2p$ electron, so the $2p$ electron is more shielded by the inner core and is held less tightly. This too is a correct statement. …