Step-by-step Solution:

1. Identify the Reaction and Reagents: The reaction involves an alkene (the prop-1-enyl group on the benzene ring) treated with diborane (B2H6) followed by hydrogen peroxide (H2O2) and a base (OH-). This is a classic two-step reaction known as hydroboration-oxidation.

2. Understand the Reaction Mechanism: Hydroboration-oxidation is a method to convert an alkene into an alcohol. The key features of this reaction are:

   • Regioselectivity: It follows anti-Markovnikov's rule. This means the hydroxyl group (-OH) adds to the less substituted carbon atom of the double bond, while the hydrogen atom adds to the more substituted carbon.
   • Stereoselectivity: It is a syn-addition, meaning the H and OH groups add to the same side of the double bond. (Stereochemistry is not a deciding factor in this particular question as only one constitutional isomer is the major product).

3. Analyze the Substrate: The starting material is p-isopropyl-(prop-1-enyl)benzene. $\text{Structure: } \quad p-(CH_3)_2CH-C_6H_4-CH=CH-CH_3$ The reaction occurs at the carbon-carbon double bond of the prop-1-enyl group (-CH=CH-CH3). The isopropyl group and the aromatic ring remain unchanged.

4. Apply Anti-Markovnikov's Rule: Let's analyze the two carbons of the double bond in the $Ar-CH=CH-CH_3$ system (where Ar is the p-isopropylphenyl group):

   • Carbon 1 (C1): The carbon directly attached to the aromatic ring (Ar-CH=). This carbon is benzylic. If a carbocation were to form here, it would be a secondary benzylic carbocation, which is highly stabilized by resonance.
   • Carbon 2 (C2): The middle carbon of the propyl chain ($=CH-CH_3$). A carbocation here would be a secondary carbocation, which is less stable than the benzylic one.

   According to the modern interpretation of Markovnikov's rule, the electrophile (like H+) adds to form the more stable carbocation. The hydroboration-oxidation gives the opposite, or anti-Markovnikov, product.

   In the hydroboration step, the boron atom (from BH3) acts as the electrophile and adds to the less substituted carbon, while the hydrogen atom adds to the more substituted carbon. The regioselectivity is governed by both steric and electronic factors:

   • Steric Factor: The borane molecule (BH3) is bulky. It preferentially attacks the less sterically hindered carbon atom. The carbon attached to the bulky phenyl group (C1) is more hindered than the carbon attached to the smaller methyl group (C2). Therefore, boron adds to C2.
   • Electronic Factor: In the transition state, a partial positive charge develops on the more substituted carbon (C1), which is stabilized by the phenyl ring. The hydrogen from borane (acting as a hydride, H⁻) adds to this carbon.

   Both factors lead to the same intermediate: the boron atom attaches to C2 and the hydrogen atom attaches to C1. $Ar-CH=CH-CH_3 \xrightarrow{1. B_2H_6} Ar-CH_2-CH(BH_2)-CH_3$

5. Oxidation Step: In the second step, the organoborane intermediate is oxidized with H2O2 in the presence of OH-. The BH2 group is replaced by an -OH group, yielding the alcohol product. $Ar-CH_2-CH(BH_2)-CH_3 \xrightarrow{2. H_2O_2, OH^-} Ar-CH_2-CH(OH)-CH_3$

6. Identify the Final Product: The final product 'X' has the structure where the original prop-1-enyl group has been converted to a 2-hydroxypropyl group. The isopropyl group at the para position is unaffected.

   The structure of 'X' is:

   This corresponds to 1-(4-isopropylphenyl)propan-2-ol.

7. Compare with Options:

   • Option A: p-(iPr)-C6H4-CH(OH)-CH2-CH3. This is the Markovnikov product. Incorrect.
   • Option B: p-(iPr)-C6H4-CH2-CH(OH)-CH3. This is the anti-Markovnikov product. Correct.
   • Option C: The hydroxyl group is on the isopropyl group. Incorrect.
   • Option D: The hydroxyl group is on the benzene ring. Incorrect.

Therefore, the major product 'X' is the structure shown in option B.