Step 1: Analyze the product $\text{X$ from Reaction (A).}
The reaction is:
\[
\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{A}} \text{X}
\]
The product $\text{X}$ dissolves in dilute $\text{HCl}$, indicating that $\text{X}$ is a base ��� most likely a primary amine ($\text{R-NH}_2$).
A nitrile ($\text{R-CN}$) converts to a primary amine ($\text{R-CH}_2\text{NH}_2$) by complete reduction:
\[
\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{A}} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 \quad (\text{X})
\]
This full reduction can occur via $\text{LiAlH}_4/\text{H}_2\text{O}$, $\text{H}_2/\text{Ni}$, or $\text{Na}/\text{Hg}, \text{C}_2\text{H}_5\text{OH}$ (Mendius reduction).
Hence, $\text{A}$ corresponds to $\text{Na}/\text{Hg}, \text{C}_2\text{H}_5\text{OH}$.
Step 2: Analyze Reaction (B).
Reaction:
\[
\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{B}} \text{Y}
\]
The product $\text{Y}$ reacts with $2,4$-DNP reagent, which detects aldehydes or ketones.
Therefore, $\text{Y}$ must be an aldehyde ($\text{R-CHO}$).
Partial reduction of a nitrile to an aldehyde is carried out by $\text{DIBAL-H}/\text{H}_2\text{O}$:
\[
\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{DIBAL-H/H}_2\text{O}} \text{C}_6\text{H}_5\text{CHO} \quad (\text{Y})
\]
Thus, $\text{B} = \text{DIBAL-H}/\text{H}_2\text{O}$.
Step 3: Final Verification.
\[
\text{A} = \text{Na}/\text{Hg}, \text{C}_2\text{H}_5\text{OH} \quad (\text{Complete reduction to amine})
\]
\[
\text{B} = \text{DIBAL-H}, \text{H}_2\text{O} \quad (\text{Partial reduction to aldehyde})
\]
Therefore, the correct option is (B).