The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is

The first-ionisation enthalpy $$I_1$$ is the energy required to remove the outer-most electron from an isolated gaseous atom.

General trends to keep in mind:
1. Along a period, $$I_1$$ usually increases because effective nuclear charge increases and atomic size decreases.
2. Down a group, $$I_1$$ usually decreases because atomic size increases and the outer electron is farther from the nucleus.
3. Exceptional dips occur when removal of an electron breaks a particularly stable electronic configuration (fully filled or half-filled subshell).

The four given elements with their valence-shell electronic configurations are:
B (Z = 5): $$2s^2\,2p^1$$
P (Z = 15): $$3s^2\,3p^3$$  (half-filled $$p$$ subshell)
S (Z = 16): $$3s^2\,3p^4$$  (one electron more than half-filled)
F (Z = 9): $$2s^2\,2p^5$$

Step-wise comparison:

• B vs F (same period, 2): Boron lies further left than fluorine, has a lower effective nuclear charge and a larger atomic radius, so $$I_1(B)$$ is much smaller than $$I_1(F)$$.
Hence $$I_1(B) \lt I_1(F)$$.

• P vs S (same period, 3): Phosphorus has a half-filled $$3p^3$$ configuration that is relatively stable. Removing an electron from sulphur breaks the paired $$3p^4$$ configuration, which already suffers inter-electronic repulsion. Therefore $$I_1(S)$$ is slightly lower than $$I_1(P)$$ despite the usual left-to-right trend.
Hence $$I_1(S) \lt I_1(P)$$.

• Comparing B with the period-3 elements: Period-3 atoms (P and S) are larger than period-2 boron, but they also possess a considerably higher effective nuclear charge. The net effect is that $$I_1(B)$$ remains the smallest of all four values.

• Fluorine is at the extreme right of period-2 with a very high effective nuclear charge and the smallest size among the four, so it possesses the highest $$I_1$$.

Collecting all inequalities:
$$I_1(B) \lt I_1(S) \lt I_1(P) \lt I_1(F)$$

Thus the increasing order (lowest first) is:
$$B \lt S \lt P \lt F$$

Option D which is: B < S < P < F