because of $\Gamma (Z,a)$. as only stack alphabet.
you have one top of stack alphabet(Z) and one simple stack alphabet(a), now you can not use Z other than top of stack, and you are left with only 'a' for comparisons,
here with 'a' as only comparing parameter-> "you can compare quantity but not quality"
now, for anbn, you need to compare only quantity i.e no. of a's = no. of b's
but for wwr or w#wr, you need to compare corresponding pairs in left half and right half i.e 'a' with 'a' and 'b' with 'b' which require comparison of both quantity and quality which cant be performed with above PDA,
so answer is option 3 only