Generally, ionization enthalpy increases across a period. However, the first ionization enthalpy of boron ($Z=5$) is slightly less than that of beryllium ($Z=4$). This is because in beryllium, the electron is removed from the $2s$ orbital ($1s^2 2s^2$), while in boron, it is removed from the $2p$ orbital ($1s^2 2s^2 2p^1$). The $2s$ electron penetrates closer to the nucleus than the $2p$ electron and is more strongly attracted. Additionally, the $2p$ electron in boron is more shielded by the inner core electrons than the $2s$ electrons of beryllium. Therefore, it is easier to remove the $2p$ electron from boron, resulting in a lower ionization energy compared to beryllium , . The only option reflecting the trend $IE_1(Be) > IE_1(B)$ is Option D.