Consider the following \( \text{NPDA} = (\{q_0, q_1, q_f\}, \{a, b\}, \{1, z\}, \delta, q_0, z,…

The transition $\delta(q_0, \lambda, z) = \{(q_f, z)\}$ allows the NPDA to accept the empty string $\lambda$ immediately (corresponding to $n = 0$). When reading the first $a$, $\delta(q_0, a, z) = \{(q_1, 11z)\}$ pushes two $1$'s onto the stack above $z$, and for each subsequent $a$, $\delta(q_1, a, 1) = \{(q_1, 111)\}$ replaces one $1$ with three $1$'s, resulting in a net increase of two $1$'s on the stack per input symbol $a$. Thus, after reading $n$ occurrences of $a$, the stack contains $2n$ symbols of $1$ above $z$. Each input symbol $b$ pops exactly one $1$ via $\delta(q_1, b, 1) = \{(q_1, \lambda)\}$, so exactly $2n$ occurrences of $b$ are required to expose $z$ and reach the accepting state $q_f$ via $\delta(q_1, \lambda, z) = \{(q_f, z)\}$. Hence, the accepted language is $L = \{a^n b^{2n} : n \ge 0\}$.