Yes, it's c. You can show that (a) = (b) as follows. Note that a + a'b = (a + a')(a + b) (by distributivity of disjunction over conjunction) = 1(a + b) = a + b. Of course, similarly a' + ab = a' + b. Using this trick and factoring out w, we get
xy' + z' + xyz + y'z =
xy' + xyz + z' + y' =
xyz + z' + y' (since xy' + y' = (x + 1)y' = 1y' = y') =
xy + z' + y' =
x + z' + y'
The fact that (b) = (d) is trivial. Nonequivalence of (c) is best verified by finding truth values where the expressions differ. Of course, equivalence can also be checked using truth tables.