Informally, whenever the PDA M sees an input a in the start state 
with the start symbol z on the top of the stack it pushes a onto the stack and changes state to 
. (to remember that it has seen the first 'a'). On state if it sees anymore a, it simply pushes it onto the stack. Note that when M is on state , the symbol on the top of the stack can only be a. On state if it sees the first b with a on the top of the stack, then it needs to start comparison of numbers of a's and b's, since all the a's at the begining of the input have already been pushed onto the stack. It start this process by popping off the a from the top of the stack and enters in state q3 (to remember that the comparison process has begun). On state 
, it expects only b's in the input (if it sees any more a in the input thus the input will not be in the proper form of anbn). Hence there is no more on input a when it is in state . On state it pops off an a from the top of the stack for every b in the input. When it sees the last b on state q3 (i.e. when the input is exaushted), then the last a from the stack will be popped off and the start symbol z is exposed. This is the only possible case when the input (i.e. on 
-input ) the PDA M will move to state 
which is an accept state.
with the start symbol z on the top of the stack it pushes a onto the stack and changes state to . (to remember that it has seen the first 'a'). On state if it sees anymore a, it simply pushes it onto the stack. Note that when M is on state , the symbol on the top of the stack can only be a. On state if it sees the first b with a on the top of the stack, then it needs to start comparison of numbers of a's and b's, since all the a's at the begining of the input have already been pushed onto the stack. It start this process by popping off the a from the top of the stack and enters in state q3 (to remember that the comparison process has begun). On state , it expects only b's in the input (if it sees any more a in the input thus the input will not be in the proper form of anbn). Hence there is no more on input a when it is in state . On state it pops off an a from the top of the stack for every b in the input. When it sees the last b on state q3 (i.e. when the input is exaushted), then the last a from the stack will be popped off and the start symbol z is exposed. This is the only possible case when the input (i.e. on -input ) the PDA M will move to state which is an accept state.