To identify the product (P) of the reaction, we need to understand the reaction mechanism.$\\$The given compound is 3-methylbutan-2-ol:$\\ CH₃–CH(CH₃)–CH(OH)–CH₃ \newline$This is a secondary alcohol. When it reacts with HBr, it undergoes an $\\ {S}_{N}1$ reaction mechanism.$\\$Steps of the reaction:$\\$1. Protonation of the alcohol group:$\\$The $$OH$$ group is protonated by HBr, forming water, a good leaving group.$\\$2. Formation of carbocation:$\\$The water molecule leaves, forming a carbocation at the second carbon.$\\$3. Nucleophilic attack by Br$^-: \\$The bromide ion attacks the carbocation, leading to the formation of the product.$\\$The carbocation is formed at the second carbon, so the bromide will attach there.$\\$The product is:$\\$CH$_3-{C}-{CH}_2-{CH}_3 \\ \phantom{{CH}_3-}\vert \\ \phantom{{CH}_3-}{CH}_3 \\ \phantom{{CH}_3-}\vert \\ \phantom{{CH}_3-}{Br} \\$This corresponds to Option 1.