The major product 'P' formed in the given reaction is

The given substrate is a substituted naphthalene containing:
$$\text{}$$
$$-CH_2CH_3$$, $$-CH=CH_2$$ and $$-CH_2COOCH_3$$
$$\text{}$$

Reaction conditions:

$$(i)\ \text{alkaline } KMnO_4,\ \Delta$$

$$(ii)\ H_3O^+$$
$$\text{}$$

Hot alkaline $$KMnO_4$$ oxidizes all benzylic side chains having at least one benzylic hydrogen into carboxylic acids.

For the ethyl group:
$$\text{}$$

$$-Ar-CH_2CH_3\rightarrow-Ar-COOH$$

For the vinyl group:

$$-Ar-CH=CH_2\rightarrow-Ar-COOH$$
$$\text{}$$

'For the ester side chain:

$$-Ar-CH_2COOCH_3\rightarrow-Ar-COOH$$
$$\text{}$$

Thus all three side chains are converted into carboxylic acid groups.

Therefore the final product is:

$${\text{Naphthalene-1,4,6-tricarboxylic acid}}$$

i.e. Option B