Find product P of the following reaction.

Correct Answer: Option 3

Let's analyze the step-by-step chemical reaction sequence starting from styrene (vinylbenzene) as shown in the question image to find the major product P.

Step 1: Acid-catalyzed hydration of Styrene
The starting material is styrene, which has a vinyl group ($-\mathrm{CH}={\mathrm{CH}}_2$) attached to a benzene ring.
When treated with dilute acid ($H_{2}O/H^{+}$), an electrophilic addition reaction occurs following Markovnikov's rule:
1. Protonation of the alkene double bond occurs to form the more stable benzylic carbocation intermediate ($C_6{H}_5-{\mathrm{CH}}^{+}-{\mathrm{CH}}_3$), which is highly stabilized by resonance with the aromatic ring.
2. Water acts as a nucleophile and attacks the carbocation, followed by deprotonation to yield a secondary alcohol: 1-phenylethanol ($C_6{H}_5-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3$).

Step 2: Oxidation with Chromic Acid
The secondary alcohol, 1-phenylethanol, is treated with chromic acid (${\mathrm{CrO}}_3/H^{+}$), which is a strong oxidizing agent.
This reagent oxidizes secondary alcohols into ketones. Therefore, 1-phenylethanol is oxidized to acetophenone ($C_6{H}_5-\mathrm{CO}-{\mathrm{CH}}_3$).

Step 3: Wolff-Kishner Reduction
Finally, acetophenone is treated with hydrazine (${\mathrm{NH}}_2-{\mathrm{NH}}_2$) in the presence of concentrated potassium hydroxide ($\mathrm{KOH}$). This is the classic Wolff-Kishner reduction:
1. Hydrazine reacts with the carbonyl group of acetophenone to form a hydrazone intermediate ($C_6{H}_5-C\left({\mathrm{CH}}_3\right)=N-{\mathrm{NH}}_2$).
2. Under strongly basic conditions with heating, the hydrazone undergoes deprotonation and elimination of nitrogen gas ($N_2$), reducing the ketone group ($-\mathrm{CO}-$) directly to a methylene group ($-{\mathrm{CH}}_2-$).
3. The resulting major product P is ethylbenzene ($C_6{H}_5-{\mathrm{CH}}_2-{\mathrm{CH}}_3$).

Comparing this structure with the options provided, Option 3 shows ethylbenzene.