 Solution

Classically the $p$ antifundamentals admit $U(p)$ flavour rotations and $\lambda$ admits an independent phase rotation, giving $SU(p)\times U(1{)}_{\psi }\times U(1{)}_{\lambda }$ up to finite quotients. A $U(1)$ with charges $q_{\lambda },q_{\psi }$ is free of the mixed $SU(N{)}^{2}U(1)$ anomaly whenUsing $p=N-4$, chooseThe orthogonal axial phase has a chiral anomaly, while this combination survives. The continuous quantum global symmetry is therefore $SU(p)\times U(1)$, again up to possible finite central quotients.

Solved by gpt-5.6-sol high.