Consider the transition diagram of a PDA given below with input alphabet $\Sigma = \{a,b\}$ and stack alphabet $\Gamma = \{X, Z\}$. $Z$ is the initial stack symbol. Let $L$ denote the language accepted by the PDA. Which one of the following is TRUE?

Let us analyze the given PDA:

1. Initial State (Start and Final State):

   • The start state is also an accepting (final) state.

   • On reading nothing ($\varepsilon$), it accepts the empty string $\varepsilon$.

   • On input $a$, it pushes $X$ onto the stack for every $a$ read ($a, Z / XZ$ and $a, X / XX$). Since this state is an accepting state, all strings consisting solely of $a$'s ($a^n$ for $n \ge 0$) are accepted by staying in the first state.

2. Transition to Second State:

   • On reading $b$ while the top of the stack is $X$, it pops an $X$ ($b, X / \varepsilon$) and moves to the second state.

   • In the second state, it continues to pop an $X$ for each subsequent $b$ read ($b, X / \varepsilon$).

3. Transition to Third State (Final State):

   • From the second state, on $\varepsilon$ input when the stack top is $Z$, it transitions to the third (accepting) state with stack unchanged ($\varepsilon, Z / Z$).

   • This path accepts strings where the number of $b$'s matches the number of $a$'s pushed, i.e., $a^n b^n$ for $n \ge 1$.

   • Including $n = 0$ (which is $\varepsilon$), all strings of the form $a^n b^n$ for $n \ge 0$ are accepted.

Combining both accepting conditions:

4. Determinism:

   • For any state, input symbol, and stack top, there is at most one valid move and no conflicting $\varepsilon$-transitions.

   • Specifically, from the first state, the transitions on $a$ and $b$ are mutually exclusive, and there is no $\varepsilon$-move from the start state.

   • Therefore, the PDA is a Deterministic Pushdown Automaton (DPDA), which means $L$ is a Deterministic Context-Free Language (DCFL).

Thus, option (D) is correct.