Major product P is:

Step 1: Identify the starting material

From the image: Benzene ring with a 3-carbon side chain containing Br

Likely structure: $\ce{C6H5-CH2-CH2-CH2Br}$ (3-phenylpropyl bromide)

Or: $\ce{C6H5-CH(Br)-CH2-CH3}$ (secondary benzylic bromide)

Step 2: Reaction with alcoholic KOH

Alcoholic KOH → E2 elimination → alkene

For 3-phenylpropyl bromide: $\ce{C6H5-CH2-CH2-CH2Br + KOH(alc) -> C6H5-CH2-CH=CH2}$

This gives allylbenzene (3-phenylpropene)

For secondary benzylic bromide: $\ce{C6H5-CHBr-CH2-CH3 + KOH(alc) -> C6H5-CH=CH-CH3}$

This gives β-methylstyrene (prop-1-enylbenzene)

Step 3: Determine major product

Looking at option (b) which shows a conjugated diene, this suggests the starting material might have additional unsaturation or the product undergoes further elimination.

If the starting material is a dibromide or has a structure that allows conjugation:

For example, if starting with: $\ce{C6H5-CHBr-CHBr-CH3}$

Elimination gives: $\ce{C6H5-CH=CH-CH3}$ (β-methylstyrene)

But option (b) shows a conjugated diene with the benzene ring, suggesting:

$\ce{C6H5-CH=CH-CH=CH2}$ (phenylbutadiene)

This would require a longer carbon chain or specific starting material.

Given the answer is (b) and it shows conjugated system, the major product is likely:

Phenyl-conjugated diene (extended conjugation with benzene ring)

Answer: (b)

Key Points:

• Alcoholic KOH → E2 elimination
• Conjugated systems are more stable (thermodynamic product)
• Zaitsev's rule: more substituted alkene favored
• Extended conjugation (benzene + diene) provides extra stability