There are two cases possible
case I) 1 couple + 2 individual or case II) 2 couples
Total sample space = total ways in which 4 person can win = 16C4 = 1820
now
case I) 1 couple + 2 individual
number of ways of choosing 1 couple = 8C1 = 8
now this two individual can be both husband or both wife or different husband and wife
both husband = 7C2 = 21 ( since one husband is already calculated in couple so 7 husband remains )
both wife = 7C2 = 21 ( same reason )
different husband and different wife = 7C1 * 6C1 = 42 ( chossing 1 husband from 7 husbands and since we cant choose the wife of choosed husband. So choosing one wife from remaining 6 wives )
So total ways in this case = 8 * ( 21+21+42 ) = 672
case II) 2 couples
number of ways of choosing 2 couples = 8C2 = 28
So probability = (672 + 28)/1820 = 5/13 = 15/39
So ans should be D)