In the given reaction scheme, P is a phenyl alkyl ether, Q is an aromatic compound; R and S are the major products.

The correct statement about S is

The correct option is It inhibits the synthesis of prostaglandin.

Step-by-step breakdown of the reaction scheme:

1. Cleavage of phenyl alkyl ether (P):
When a phenyl alkyl ether P (such as anisole, C₆H₅-O-R) reacts with hydrogen iodide (HI), cleavage of the ether bond occurs to form phenol as the aromatic product Q.

$\text{P (Phenyl alkyl ether)}+\text{HI}\to \text{Q (Phenol,}{\text{C}}_6{\text{H}}_5\text{OH})+\text{Alkyl iodide}$

2. Kolbe-Schmitt Reaction of Phenol (Q):
Phenol (Q) is treated with:
(i) NaOH to form sodium phenoxide
(ii) CO₂ as an electrophile
(iii) H₃O⁺ for acidification
This is the classic Kolbe's reaction, yielding salicylic acid (2-hydroxybenzoic acid) as the major product R.

$\text{Q (Phenol)}\underset{\left(\text{iii}\right){\text{H}}_3{\text{O}}^{+}}{\overset{\left(\text{i}\right)\text{NaOH,}\left(\text{ii}\right){\text{CO}}_2}{\to }}\text{R (Salicylic acid)}$

3. Acetylation of Salicylic Acid (R):
Treating salicylic acid (R) with acetic anhydride ((CH₃CO)₂O) in acidic medium results in acetylation of the phenolic -OH group to yield acetylsalicylic acid, commonly known as aspirin (S).

$\text{R (Salicylic acid)}\underset{\left(\text{ii}\right){\text{H}}_3{\text{O}}^{+}}{\overset{\left(\text{i}\right){\left({\text{CH}}_3\text{CO}\right)}_2\text{O}}{\to }}\text{S (Aspirin / 2-acetoxybenzoic acid)}$

Properties and Action of Aspirin (S):
Aspirin is a well-known non-narcotic analgesic and antipyretic drug. Its mode of action involves the inhibition of the cyclooxygenase (COX) enzyme, which prevents the synthesis of prostaglandins that cause inflammation, fever, and pain.
Therefore, the correct statement regarding compound S is that it inhibits the synthesis of prostaglandin.