Step 1: Reaction of 1-phenylethanol with HBr.
The given compound is
\[
C_6H_5CH(OH)CH_3
\]
On reaction with HBr, the \(-OH\) group is replaced by bromine.
So, the product formed is
\[
C_6H_5CHBrCH_3
\]
Step 2: Reaction with alcoholic KOH and heat.
Alcoholic KOH causes dehydrohalogenation.
Thus,
\[
C_6H_5CHBrCH_3
\]
loses HBr to form styrene:
\[
C_6H_5CH=CH_2
\]
Therefore,
\[
A=\text{styrene}
\]
Step 3: Hydroboration oxidation of styrene.
Styrene undergoes hydroboration oxidation with
\[
B_2H_6
\]
followed by
\[
NaOH/H_2O_2
\]
This reaction gives anti-Markovnikov addition of water.
Hence, the \(-OH\) group attaches to the terminal carbon of the double bond.
So,
\[
C_6H_5CH=CH_2
\longrightarrow
C_6H_5CH_2CH_2OH
\]
Step 4: Final conclusion.
The major product \(B\) is
\[
\boxed{C_6H_5CH_2CH_2OH}
\]
which is 2-phenylethanol.
Hence, the correct option is
\[
\boxed{(4)}
\]