According to Varadarajan$^{1}$, in a separable Hilbert space,
If $A$ is of trace class and $B$ is any bounded operator, $AB$ and $BA$ are of trace class; and moreover, $tr(AB) = tr(BA)$.
Orthogonal projections are bounded, since $\|Px\|\le \|x\|$, so if $D$ is of trace class, then according to the above statement, $DP$ is also trace class and $\mathrm{tr}(P(DP))=\mathrm{tr}((DP)P)=\mathrm{tr}(D(PP))=\mathrm{tr}(DP)$ since $P^2=P$.
However, I have some doubts because Varadarajan consistently writes $\mathrm{tr}(PDP)$ without this simplification. Where am I wrong?
$^{1}$Varadarajan, Geometry of Quantum Theory Springer 1985, Second edition, p. 84