The product formed in the following multistep reaction is:

Following the reaction sequence with propene, the first stage involves hydroboration with $$B_2H_6$$, leading to anti‐Markovnikov addition of $$BH_3$$.
Subsequent oxidation with $$H_2O_2, NaOH$$ converts the intermediate to the anti‐Markovnikov alcohol $$CH_3CH_2CH_2OH$$ (1-propanol).
The primary alcohol is then oxidized by pyridinium chlorochromate ($$PCC$$) to afford the aldehyde $$CH_3CH_2CHO$$ (propanal).
In the final step, reaction with the Grignard reagent $$CH_3MgBr$$ followed by hydrolysis yields the secondary alcohol $$CH_3CH_2CH(OH)CH_3$$ (2-butanol).

The overall product is $$CH_3CH_2CH(OH)CH_3$$ (butan-2-ol), which corresponds to option 3.