Consider the following reaction sequence. The product 'B' is

The starting compound is:

$$\mathrm{p\text{-}Aminobenzonitrile}$$

It reacts with:

$$\mathrm{DIBAL\text{-}H}$$

followed by hydrolysis.

DIBAL-H selectively reduces the nitrile group:

$$\mathrm{-CN}$$

to an aldehyde.

$$\mathrm{H_2N-C_6H_4-CN \xrightarrow[DIBAL\text{-}H]{H_2O} H_2N-C_6H_4-CHO}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{p\text{-}Aminobenzaldehyde}$$

Compound $$\mathrm{A}$$ then reacts with acetaldehyde:

$$\mathrm{CH_3CHO}$$

in presence of:

$$\mathrm{NaOH/\Delta}$$

This is a:

$$\mathrm{Cross\ Aldol\ Condensation}$$

Compound $$\mathrm{A}$$ has no:

$$\mathrm{\alpha\text{-}hydrogen}$$

Hence, it acts only as the electrophile.

Acetaldehyde possesses:

$$\mathrm{\alpha\text{-}hydrogens}$$

and forms an enolate ion.

The enolate attacks the carbonyl carbon of compound $$\mathrm{A}$$.

Subsequent dehydration forms an:

$$\mathrm{\alpha,\beta\text{-}unsaturated\ aldehyde}$$

The final product $$\mathrm{B}$$ is:

$$\mathrm{H_2N-C_6H_4-CH=CH-CHO}$$

Correct option:

$$\mathrm{B}$$