I can't understand this example of PDA Automata
The idea of this PDA is very simple. Start with a PDA for $L' = \{ w \in \{a,b\}^* : |w|_a = |w|_b \}$, and now modify it as follows:


Whenever you read $a$, simulate the original PDA reading $aa$.
Whenever you read $b$, simulate the original PDA reading $bbb$.


Hopefully you can see why the new PDA accepts $L$ given that the original one accepted $L'$.

The PDA for $L'$ maintains a stack which can be of three different types, depending on the word $x$ seen so far:


If $|x|_a = |x|_b$ then the stack contents is $Z_0$. In particular, this is also the initial stack.
If $|x|_a &gt; |x|_b$ then the stack contents is $B^{|x|_a - |x|_b} Z_0$.
If $|x|_a &lt; |x|_b$ then the stack contents is $A^{|x|_b - |x|_a} Z_0$.


When the top-of-stack is $Z_0$ we know that $|x|_a = |x|_b$, and so we optionally move to the final state; the PDA will accept only if $x = w$, since otherwise it will get stuck.

When converting this PDA to the one for $L$, we use epsilon transitions to simulate reading the second $a$ and the second and third $b$s.