1. Identify compound $P$

   The statement says that in compound $P$, $Et$ denotes an ethyl group. This standard question corresponds to an ester of benzoic acid, i.e. ethyl benzoate: $P = C_6H_5COOEt$

2. Reaction of $P$ with excess $CH_3MgBr$ followed by hydrolysis

   Esters react with excess Grignard reagent with two additions of the alkyl group.

   For ethyl benzoate: $C_6H_5COOEt \xrightarrow[ether]{2CH_3MgBr} C_6H_5C(OMgBr)(CH_3)_2 \xrightarrow{H_2O} C_6H_5C(OH)(CH_3)_2$

   Hence, $Q = C_6H_5C(OH)(CH_3)_2$ which is 2-phenyl-2-propanol (cumyl alcohol), a tertiary alcohol.

3. Treatment of $Q$ with $H_2SO_4$ at $0^\circ C$

   Tertiary benzylic alcohol undergoes dehydration readily to form the corresponding alkene: $C_6H_5C(OH)(CH_3)_2 \xrightarrow{H_2SO_4} C_6H_5C(CH_3)=CH_2$

   Therefore, $R = \alpha\text{-methylstyrene} = \text{isopropenyl benzene}$

4. Reaction of $R$ with $CH_3COCl/AlCl_3$

   This is a Friedel–Crafts acylation. The substituent on benzene is: $-C(CH_3)=CH_2$

   This alkenyl side chain behaves like an alkyl-type activating group through hyperconjugative/electron-releasing effect and directs incoming electrophile to the ortho/para positions. Because of steric hindrance, the para product predominates.

   Thus acetyl group $(-COCH_3)$ enters mainly at the para position relative to the isopropenyl group.

   So, $S = p\text{-acetyl-}\alpha\text{-methylstyrene}$

   Structure: $\boxed{\text{para-}CH_3CO\text{-}C_6H_4\text{-}C(CH_3)=CH_2}$

5. Matching with options

   The option corresponding to para-acetyl-$\alpha$-methylstyrene is Option A.

Final Answer

$\boxed{A}$