imo b
codes are as follows,try adding one at a time to find minimum necessary
1 a
2 b
3 ab
4 c
5 ac
6 bc
7 d
8 ad
9 bd
10 cd
11
12
hence using a,b,c,d 4 letters we can create 10 distinct codes,so 5 minimum are necessary.

gmatguy16, the approach u followed is really appreaciatable but I took a different approach -
Now if there are n tokens, taking 2 at a time to indicate a different unique combination - the combination will be = (n - 1)!. Are you getting me, showoff16884? For more manual clarification, just refer gmatguy16's approach to take 2 tokens at a time.
Now if n = 4, (4 -1)! = 6
So total representation = two digit representation + single digit representation = 6 + 4 = 10 -> NOT SUFF for 12 people.
So the least no of alphabets to chose is = 5. Am I clear to explain?

Thank you gmatguy & amitava...B it is!!!