The major product P formed in the given reaction is:

Step 1: Identify the Starting Material

From the image: a naphthalene ring with $\ce{-CH2CH3}$ (ethyl), $\ce{-CH=CH2}$ (vinyl), and $\ce{-COOCH3}$ (methyl ester) substituents.

Step 2: Hot Alkaline KMnO4 Oxidation

Hot alkaline $\ce{KMnO4}$ oxidizes:

• Alkyl side chains on benzene ring to $\ce{-COOH}$
• Vinyl group $\ce{-CH=CH2}$ to $\ce{-COOH}$ (oxidative cleavage)
• Methyl ester $\ce{-COOCH3}$ hydrolyzes to $\ce{-COOH}$ under alk. conditions
• The naphthalene ring itself may be cleaved to give multiple $\ce{-COOH}$ groups

Step 3: Product P All oxidizable side chains $\rightarrow$ $\ce{-COOH}$. The complex oxidation gives a phthalic acid or naphthalenedicarboxylic acid type product (option d from the image).

Key Points:

• Hot alk. $\ce{KMnO4}$: oxidizes all alkyl side chains to $\ce{-COOH}$
• $\ce{-CH=CH2}$ oxidatively cleaved to $\ce{-COOH}$
• Ester groups hydrolyze under alkaline conditions
• Naphthalene ring also susceptible to oxidation under harsh conditions