Consider the following sequence of reactions

Compound '$$C$$' is

The starting material is propene, $$CH_3CH=CH_2$$.

Step (i) : Hydroboration-oxidation
Reagents $$BH_3$$ / $$H_2O_2$$ / $$OH^-$$ add $$H$$ and $$OH$$ across the double bond in an anti-Markovnikov manner with syn stereochemistry.
Hence the -OH group attaches to the terminal carbon:$$CH_3CH=CH_2 \;\xrightarrow[\;H_2O_2/OH^-\;]{\;BH_3\;} CH_3CH_2CH_2OH$$
Thus $$A$$ is 1-propanol.

Step (ii) : PCC oxidation
Pyridinium chlorochromate (PCC) oxidises a primary alcohol only up to the aldehyde stage:$$CH_3CH_2CH_2OH \;\xrightarrow{\;PCC\;} CH_3CH_2CHO$$
Hence $$B$$ is propanal.

Step (iii) : Grignard addition followed by acidic work-up
A Grignard reagent adds the alkyl group to the carbonyl carbon and converts the $$C=O$$ to $$C-OH$$ after hydrolysis:$$CH_3CH_2CHO + CH_3MgBr \;\xrightarrow[\;H_2O\;]{} CH_3CH_2CH(OH)CH_3$$
The carbonyl carbon of propanal gains a $$CH_3$$ group, giving a secondary alcohol, 2-butanol.

Therefore compound $$C$$ is $$CH_3CH_2CH(OH)CH_3$$, i.e. 2-butanol.

Option A which is: $$CH_3CH_2CH(OH)CH_3$$