Answer:
Option C
Explanation:
Only -CHO group is to be reduced to -CH2OH
It can be done using $NaBH_{4}$ in $C_{2}H_{5}OH$
(a)$LiAlH_{4} / (C_{2}H_{5})_{2}O$ reduces I,II and III into - $CH_{2}OH$ , and IV into diol.
(b) $BH_{3}$ / THF show same properties as (a)
(c) $NaBH_{4}$ / $C_{2}H_{5}OH$ reduces III into -$CH_{2}OH$
(d) Raney nickel, same as (a) and (b) thus (c) is correct reagent