Working with PDA transitions

Question

Either I dont understand PDA at all or this question is wrong or I am missing something very basic:

Comments

state q0 denotes even no of a's and state q1 denotes odd no of a's.

Here we consider even no of a's then b and c should be equall.

at state q1 (c,b/$\epsilon$)  //pop 'b' for one 'c'.

at state q2 also same pop 'b' for one 'c'.

If a's are odd then no restriction on a's and b's.

at state q4 'b's are pushed into the stack.

when (c,b/cb)  push 'c' into stack or (c,b/b) skip 'b's.

same as q5 also.

Hence W=cb or b,X=$\epsilon$,Y=$\epsilon$,Z=cb or b

Answer 1

Case1: if m is even then n=k number of b’s = number of c’s. Therefore X and Y both must represent pop operation.

Case2: if m is odd then no constraint number of b’s,number of c’s just that c must come after b. from state Q5 to Qf there is move for which b must be in the stack, so we cannot push c into stack and must ignore c. Therefore W and Z both must represent ignore operation i.e do nothing with stack.

Hence, Option B is correct.