1. Identify the compounds and the reaction trend

In nitration of aromatic amines/anilides, the rate depends on how strongly the substituent donates electron density to the benzene ring.

The usual order of activating effect is:

$-NHCOCH_3 > -NHC_6H_5 > -NH_2^+ \,\text{(very deactivating)}$

and free aniline in strongly acidic nitrating mixture is largely protonated to anilinium ion, so its ring becomes much less reactive.

2. Key principle for each compound

• Aniline in nitrating mixture gets protonated: $C_6H_5NH_2 \rightarrow C_6H_5NH_3^+$ This makes it least reactive toward electrophilic nitration.

• Diphenylamine $((C_6H_5)_2NH)$ has the nitrogen lone pair shared with two phenyl rings, so donation to any one ring is less than in acetanilide, but it is still more activating than protonated aniline.

• Acetanilide $C_6H_5NHCOCH_3$ has the group $-NHCOCH_3$, which still donates by resonance to the ring and avoids strong protonation problems of aniline under nitration conditions. Hence it nitrates more easily than aniline and diphenylamine.

• N,N-dimethylaniline (if present among the figures) has a strongly electron-donating amino group, making the ring most activated toward nitration among these protected/unprotonation-comparable cases.

3. Relative reactivity

Thus the increasing order is:

$\text{aniline} < \text{diphenylamine} < \text{acetanilide} < \text{N,N-dialkylaniline}$

which corresponds to

$(a) < (b) < (d) < (c)$

4. Match with options

This is Option D.

Final Answer

$\boxed{(a) < (b) < (d) < (c)}$

So, the correct option is D.