What language is accepted by the pushdown automaton

$M=({q_0,q_1,q_2},\{a,b\},\{a,b,z\},\delta,q_0,z,\{q_2\})$

with

$\delta(q_0,a,a)=\{(q_0,aa)\}$

$\delta(q_0,b,a)=\{(q_0,ba)\}$

$\delta(q_0,a,b)=\{(q_0,ab)\}$

$\delta(q_0,b,b)=\{(q_0,bb)\}$

$\delta(q_0,a,z)=\{(q_0,az)\}$

$\delta(q_0,b,z)=\{(q_0,bz)\}$

$\delta(q_0,\lambda,b)=\{(q_1,b)\}$

$\delta(q_0,\lambda,a)=\{(q_1,a)\}$

$\delta(q_1,a,a)=\{(q_1,\lambda)\}$

$\delta(q_1,b,b)=\{(q_1,\lambda)\}$

$\delta(q_1,\lambda,z)=\{(q_2,z)\}$

In state $q_0$, the PDA reads input symbols and pushes them into the stack.

So, first part of the string is stored in stack.

After that, using $\lambda$ move, PDA goes from $q_0$ to $q_1$.

In state $q_1$, the PDA reads input and matches it with the top of stack.

If input is $a$ and stack top is $a$, it pops $a$.

If input is $b$ and stack top is $b$, it pops $b$.

So, second part of the string must be reverse of the first part.

When stack reaches bottom symbol $z$, PDA goes to final state $q_2$.

Therefore, language accepted is even length palindrome over $\{a,b\}$.

So,

$L=\{ww^R\mid w\in\{a,b\}^+\}$