Step 1: The reaction of \( \ce{CH3CH2CH2OH} \) with \( \ce{PBr3} \) gives \( \ce{CH3CH2CH2Br} \) — a substitution reaction replacing –OH with –Br.
Step 2: The reaction of propene (\( \ce{CH3CH=CH2} \)) with HBr in the presence of benzoyl peroxide (\( \ce{(C6H5COO)2} \)) proceeds via anti-Markovnikov addition due to free radical mechanism, yielding \( \ce{CH2BrCH2CH3} \) as major product.
Step 3: So, X is \( \ce{CH3CH2CH2Br} \) and Y is \( \ce{CH2BrCH2CH3} \), which corresponds to option (3).