Take an FD, e.g. AB→C
Augment until all attributes are mentioned, e.g. ABDEFG → CDEFG (note that this is equivalent to ABDEFG → ABCDEFG because it is trivially true that A->A and B->B).
This tells you that ABDEFG is a superkey.
Check the other FDs in which the LHS is a subset of your superkey, and that on its RHS contains some other attribute of your superkey.
There are two. EF→G and FG→E. Remove the attributes of the RHS of these from your superkey. Doing so gives you two keys, that are certainly superkeys, but not necessarily irreducible ones: ABDEF and ABDFG.
However, from AB→C and CD→E we can also derive that ABD→E. Hence we can also remove the E from our ABDEF key. The nasty thing here is that when I said “Check the other FDs”, that apparently means that you should also check any FD that appears in the closure of your set of FDs (i.e. any FD that is derivable from your given set of FDs)… And that’s a bit impractical to do by hand …
A useful site for verifying whether your results are correct:
http://raymondcho.net/RelationalDatabaseTools/
You should now be able to determine that option c is a key.