= Solution
Without quencher, $\phi_{P,0}=k_P/(k_P+k_D)$. Taking the ratio gives the <Stern-Volmer equation>
$$
\boxed{\frac{\phi_{P,0}}{\phi_P}=1+\frac{k_Q}{k_P+k_D}[Q]
=1+k_Q\tau_0[Q],\qquad\tau_0=(k_P+k_D)^{-1}.}
$$
Thus the plot has \b[unit intercept and slope $k_Q\tau_0$]. A triplet formation yield less than one would cancel from this ratio if it remained independent of quencher concentration. The result assumes dynamic triplet quenching and no additional quencher-induced absorption or singlet loss.<Stern-Volmer kinetic relationships in the IUPAC Gold Book>[Definition and dynamic-quenching relation between quantum-yield ratios, quencher concentration and unquenched lifetime]{ref}
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