Solution

The language accepted by the NPDA is L = {a²ⁿbⁿ ∶ n ≥ 0}.

Explanation:

• The NPDA starts in the initial state q0 with the stack symbol z. When it reads the first symbol a, it pushes two 1's on the stack (δ(q0, a, z) = {(q1, 11z)}). Then, every time it reads another a, it pushes another 1 on the stack (δ(q1, a, 1) = {(q1, 111)}). This way, the stack stores the number of a's that have been read so far.
• When the NPDA reads a b, it pops one 1 from the stack (δ(q1, b, 1) = {(q1, λ)}). When the stack becomes empty (i.e., when all the a's have been paired with b's), the NPDA transitions to the final state qf (δ(q1, λ, z) = {(qf, z)}).
• Therefore, the NPDA accepts the language L = {a²ⁿbⁿ ∶ n ≥ 0}, which consists of strings of a's followed by an equal number of b's. The other options are not accepted by the NPDA.