Step 1 - Translate $pK_b$ into basic strength.
$$pK_b = -\log K_b$$
LOWEST $pK_b$ means HIGHEST $K_b$, i.e. the STRONGEST BASE. Basic strength depends on
the availability of the lone pair on nitrogen: the more electron density on N,
the stronger the base.

Step 2 - Examine p-nitroaniline, option (D).
The $-NO_2$ group is strongly electron-withdrawing by both $-I$ and $-R$ effects, and
being at the para position it withdraws the nitrogen lone pair very effectively.
This is the WEAKEST base, so it has the HIGHEST $pK_b$.

Step 3 - Examine aniline, option (C).
In $C_6H_5NH_2$ the lone pair on N is delocalised into the benzene ring by resonance,
so it is much less available. Aniline is a weak base ($pK_b \approx 9.38$).

Step 4 - Compare the N-alkylated anilines, options (A) and (B).
Every methyl group attached to nitrogen pushes electron density towards N by the
$+I$ (electron releasing) inductive effect, which partly offsets the resonance
withdrawal by the ring.
$C_6H_5NH(CH_3)$ has ONE methyl group ($pK_b \approx 9.30$).
$C_6H_5N(CH_3)_2$ has TWO methyl groups ($pK_b \approx 8.92$), so it has the greatest
electron density on nitrogen among the four.

Step 5 - Arrange in order.
Basic strength: $C_6H_5N(CH_3)_2 > C_6H_5NH(CH_3) > C_6H_5NH_2 > p\text{-}O_2N\text{-}C_6H_4\text{-}NH_2$
Therefore $pK_b$ : $C_6H_5N(CH_3)_2$ is the LOWEST.

Answer: (A) $C_6H_5-N(CH_3)_2$