The correct option is C Both can be distinguished by victor meyer's test
CH2=CH−CH3i. Hg(OAc)2, THF−−−−−−−−−−−→ii. NaBH4CH3−CH(OH)−CH3
CH2=CH−CH3i. B2H6, THF−−−−−−−−−→ii. H2O2, OH−(OH)CH2−CH2−CH3
P is secondary alcohol and Q is primary alcohol. Hence they can be distinguished by Victor meyer's test.