Step-by-Step Solution:

1. Analyze the Starting Material: The starting molecule is tris(3-isopropyl-5-methylphenyl)methane. This means a central carbon atom is bonded to a hydrogen and three identical substituted phenyl rings. Each phenyl ring has a methyl group (-CH3) and an isopropyl group (-CH(CH3)2) at positions meta to the point of attachment to the central carbon.

2. Analyze the First Reaction: Oxidation with hot KMnO4

   • Hot, concentrated potassium permanganate (KMnO4) is a powerful oxidizing agent used to oxidize alkyl side chains on a benzene ring.
   • Any alkyl group with at least one benzylic hydrogen (a hydrogen on the carbon atom directly attached to the benzene ring) is oxidized to a carboxylic acid group (-COOH).
   • In our starting material, each of the three phenyl rings has two alkyl side chains:
     • A methyl group (-CH3): The benzylic carbon has 3 hydrogens. It will be oxidized to -COOH.
     • An isopropyl group (-CH(CH3)2): The benzylic carbon has 1 hydrogen. It will also be oxidized to -COOH.
   • Thus, for each of the three rings, both the methyl and isopropyl groups are converted to carboxylic acid groups. The central C-H group is sterically hindered and less likely to react, so we consider the oxidation of the side chains as the major reaction pathway.
   • The intermediate product formed is tris(3,5-dicarboxyphenyl)methane.
   • This intermediate molecule has 3 rings, each with 2 carboxylic acid groups. Total number of -COOH groups = 3 * 2 = 6.

   Structure after Step 1: $\text{CH(C}_6\text{H}_3\text{(COOH)}_2)_3$

3. Analyze the Second Reaction: Reaction with SOCl2

   • Thionyl chloride (SOCl2) is a standard reagent for converting carboxylic acids (-COOH) into acid chlorides (-COCl).
   • All 6 carboxylic acid groups in the intermediate will be converted to acid chloride groups.
   • R-COOH + SOCl2 -> R-COCl + SO2 + HCl
   • The product formed is tris(3,5-bis(chlorocarbonyl)phenyl)methane.
   • This molecule contains 6 acid chloride (-COCl) groups.

   Structure after Step 2: $\text{CH(C}_6\text{H}_3\text{(COCl)}_2)_3$

4. Analyze the Third Reaction: Reaction with CH3MgBr (excess) followed by H3O+

   • This is a Grignard reaction. Methylmagnesium bromide (CH3MgBr) is a strong nucleophile.
   • Acid chlorides react with two equivalents of a Grignard reagent to form a tertiary alcohol after acidic workup.
   • The reaction for each -COCl group proceeds as follows:
     1. Nucleophilic attack by CH3- on the carbonyl carbon of the acid chloride forms a ketone: R-COCl + CH3MgBr -> R-CO-CH3 + MgBrCl.
     2. The resulting ketone immediately reacts with a second molecule of CH3MgBr: R-CO-CH3 + CH3MgBr -> R-C(OMgBr)(CH3)2.
     3. Acidic workup (H3O+) protonates the alkoxide to form the tertiary alcohol: R-C(OMgBr)(CH3)2 + H3O+ -> R-C(OH)(CH3)2.
   • Since there are 6 -COCl groups in the molecule and CH3MgBr is used in excess, all 6 groups will undergo this transformation.
   • Each -COCl group is converted into a 2-hydroxyprop-2-yl group (-C(OH)(CH3)2).

   Structure of the Final Product P: $\text{CH(C}_6\text{H}_3\text{[C(OH)(CH}_3)_2]_2)_3$

5. Count the Number of Hydroxyl Groups in Product P

   • The final product P contains 6 of the -C(OH)(CH3)2 groups.
   • Each of these groups contains one hydroxyl (-OH) group.
   • Therefore, the total number of hydroxyl groups in one molecule of the major product P is 6.

Final Answer: The total number of hydroxyl groups is 6.