The major product P of the given reaction is (where, Me is $$-$$CH$$_3$$)

In the presence of acid, the hydroxyl group is first protonated to convert it into a good leaving group.

$$R-OH + H^+ \rightarrow R-OH_2^+$$

The protonated alcohol then loses a water molecule to form a secondary carbocation.

$$R-OH_2^+ \rightarrow R^+ + H_2O$$

The secondary carbocation undergoes a $$1,2$$-methyl shift because it is adjacent to a quaternary carbon. Migration of one methyl group generates a more stable tertiary carbocation.

Carbocation rearrangement:

$$2^\circ\ \text{carbocation} \xrightarrow{\text{1,2-methyl shift}} 3^\circ\ \text{carbocation}$$

The tertiary carbocation then undergoes elimination. A base removes a proton from an adjacent carbon, leading to the formation of an alkene.

According to Zaitsev's rule, the major product is the most substituted and therefore the most stable alkene.

Removal of the proton from the carbon bearing the migrated methyl group gives a tetrasubstituted alkene, which is the most stable product.

Hence, the major product is the rearranged alkene having the double bond between the two most substituted carbon atoms.

Therefore, the correct answer is the structure shown in Option (C).