There are two condition for a function being self dual.
1- it should be neutral function. (no. of minter = no. of max term)
2-no mutually two exclusive term should be there like $($$0-7$ are mutually exclusive $1-6, 2-5, 3-4$$)$ from these pairs only one should be there.
Clearly, there are $4$ minterm, so number of minterms = no. of maxterms.
and second condition is also satisfied. So, it is a self dual function .
b) excess-3 to BCD
@Tendua awesome explanation but just for the sake of correction POS is asked and not SOP.
checking self dual of a function.
f=Σ(0,1,2,4) = a'b'c' + a'b'c + a'bc' + ab'c'
now write it in max term form (a'+b'+c')(a'+b'+c)(a'+b+c')(a+b'+c')
7 6 5 3
=$\prod$(3,5,6,7) =$\sum$(0,1,2,4) = given function so its a self dual
(b) Excess-3 to BCD
Minimal product of sum (b4, b3, b2, b1)
a) It is a self dual function, because it satisfies both conditions of self dual fuction
->It should be neutral (i.e no. of minterms is equal to the no. of maxterms)
->The function does not contain two mutually exclusive term ( eg ABC mutually exclusive to A'B'C' , AB'C' mutually excluseive to A'BC)
b) Excess-3 to BCD converter