The first ionization enthalpies of Be, B, N and O follow the order:

🧠 Two anomalies in this set: Be > B (filled $2s^2$) and N > O (half-filled $2p^3$) Across Period 2, $IE_1$ rises overall, but two breaks interrupt it. Group 2 (Be) is higher than Group 13 (B) because removing an electron from filled $2s^2$ is harder than from a lone $2p^1$. Group 15 (N) is higher than Group 16 (O) because half-filled $2p^3$ has extra symmetry stability that beats the paired-electron repulsion in $2p^4$.

🗺️ Build the order B has the lone $2p^1$ — easiest to remove. Be has filled $2s^2$ — harder than B. O has $2p^4$ with one paired orbital — pair repulsion makes it easier than N. N has half-filled $2p^3$ — hardest in this set.

So: $\ce{B} < \ce{Be} < \ce{O} < \ce{N}$.

⚠️ The trap Option (d) $\ce{B} < \ce{Be} < \ce{N} < \ce{O}$ catches the Be–B anomaly but forgets the N–O one. It places O above N, following the simple "$IE$ rises across" rule. Wrong — the half-filled $2p^3$ in N is more stable than the $2p^4$ of O, so N is harder to ionize. Both anomalies must be applied at the same time.

$\boxed{\text{Answer: (a)}}$