The increasing order of the reactivity of the following with LiAlH$$_4$$ is:

LiAlH4 is a strong nucleophilic reducing agent that attacks the carbonyl carbon of carboxylic-acid derivatives by the nucleophilic acyl-substitution mechanism.

For such reactions, the rate depends mainly on the quality of the leaving group $$Y^-$$ in the tetrahedral intermediate: better leaving groups give faster reactions. The general order of leaving-group ability (best → worst) is: $$Cl^- \gt RCOO^- \gt RO^- \gt NH_2^-$$.

Matching each derivative with its leaving group:

$$C_2H_5COCl$$ (acid chloride, C)  →  leaves $$Cl^-$$ (best)

$$(C_2H_5CO)_2O$$ (acid anhydride, D)  →  leaves $$RCOO^-$$

$$C_2H_5COOCH_3$$ (ester, B)  →  leaves $$RO^-$$

$$C_2H_5CONH_2$$ (amide, A)  →  leaves $$NH_2^-$$ (worst)

Therefore the reactivity order toward LiAlH4 is

$$\text{amide (A)} \lt \text{ester (B)} \lt \text{anhydride (D)} \lt \text{acid chloride (C)}$$

Hence, the increasing order asked in the question is: (A) < (B) < (D) < (C).

Option C matches this sequence.

Final answer: Option C