= Solution
For <radiative temperature uncertainty after background subtraction>, write $P_{\rm back}=B$, so $P_{\rm total}\simeq10B$ and $P\simeq9B$. The two absolute uncertainties are both $0.1B$. If they are independent <standard uncertainties>,
$$
\frac{\sigma_T}{T}=\frac14\frac{\sigma_P}{P}
=\frac{\sqrt{(0.1B)^2+(0.1B)^2}}{36B}
=\boxed{3.93\times10^{-3}\simeq0.39\%}.
$$
If the percentages instead denote absolute error bounds, linear worst-case propagation gives \b[0.56%]. With <correlation coefficient> $\rho$, the standard-uncertainty result is $\sqrt{2(1-\rho)}/360$; subtraction makes positive common errors partly cancel. The independence convention is therefore an assumption, not information supplied by the percentages.
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