= Solution
The <steady-state approximation> for the excited singlet gives
$$
0=I_{\rm abs}-(k_F+k_{\rm ISC})[S_1],
$$
and that for the triplet gives
$$
0=k_{\rm ISC}[S_1]-(k_P+k_D+k_Q[Q])[T_1].
$$
Combining these with the <quantum yield> definition gives the more general expression
$$
\phi_P=\frac{k_{\rm ISC}}{k_F+k_{\rm ISC}}
\frac{k_P}{k_P+k_D+k_Q[Q]}.
$$
Under the supplied limit $k_{\rm ISC}\gg k_F$, this becomes
$$
\boxed{\phi_P\simeq\frac{k_P}{k_P+k_D+k_Q[Q]}.}
$$
The approximation is the near-unit triplet formation yield, not an assumption that phosphorescence is the only triplet loss process.
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