 Solution

Propagate each output observable backwards through the Clifford circuit and writeThe $Q_i$ are mutually commuting Pauli operators, and measuring them on $|0{⟩}^{\otimes n}\otimes |\varphi ⟩$ has exactly the required joint output distribution. Initially the first $n$ qubits are constrained by the stabilizer generators $Z_1,\dots ,Z_n$. Process the $Q_i$ in order. If $Q_i$ anticommutes with a current generator, part b gives a uniform outcome and a Clifford operation that replaces that generator by $Q_i$; this step needs no measurement on $|\varphi ⟩$. If $Q_i$ commutes with every current generator, multiply it by known generators to remove its action on the first register. What remains is a Pauli observable $P_j$ on the $t$ resource qubits and is measured there. The nontrivial $P_j$ are independent and mutually commuting. An independent commuting family of Pauli operators on $t$ qubits has at most $t$ members, so $s\le t$. All effective observables are fixed by the original commuting family; the sampled outcomes merely update the classical Clifford frame. The resulting nonadaptive Pauli-based computation, followed by the stated $Z_1,\dots ,Z_n$ outputs, is therefore a weak classical simulation of a quantum circuit with the same joint distribution.