The major product B formed in the following reaction sequence is:

The starting compound is a primary alcohol. PCC (pyridinium chlorochromate) is a mild oxidising agent that stops at the aldehyde stage.

Step-1 (action of PCC)
$$C_6H_5CH_2CH_2OH \xrightarrow[\text{dichloromethane}]{\text{PCC}} C_6H_5CH_2CHO \;$$ $$-(1)$$
Thus compound $$A$$ is $$C_6H_5CH_2CHO$$ (phenyl-acetaldehyde).

Step-2 (action of hot, alkaline $$KMnO_4$$ followed by acidification)
Alkaline $$KMnO_4$$ is a strong oxidising agent. Any side chain attached to an aromatic ring that still contains at least one benzylic hydrogen is oxidised completely to a carboxyl group $$\bigl(-COOH\bigr)$$.

In $$A$$ the benzylic carbon is the one bearing the aldehydic $$-CHO$$ group in $$C_6H_5CH_2\underline{CHO}$$. It possesses benzylic hydrogens, so the entire -CH₂CHO side chain is shortened and converted to -COOH:

$$C_6H_5CH_2CHO \xrightarrow[\Delta]{KMnO_4/OH^-} C_6H_5COO^-K^+ \xrightarrow{H^+} C_6H_5COOH \;$$ $$-(2)$$

Hence compound $$B$$ is benzoic acid, $$C_6H_5COOH$$.

Among the given structures, benzoic acid corresponds to Option D.

Therefore, the major product $$B$$ is obtained as benzoic acid.
Answer: Option D which is: benzoic acid $$\bigl(C_6H_5COOH\bigr)$$.