Let $\mathcal{H}$ be a Hilbert space. It seems obvious at first glace if $T: \mathcal{H} \to \mathcal{H}$ is bounded, self-adjoint operator and $P: \mathcal{H} \to M \subseteq \mathcal{H}$ is an orthogonal projection on a closed subspace of $\mathcal{H}$, then we should have
$$ \langle PTPx,y\rangle = \langle TPx,Py \rangle = \langle Px,TPy \rangle = \langle x,PTPy \rangle. $$ Am I overthinking this, or is there some counter-example I didn't notice?
Your suspicion is not unfounded: indeed, it is not true in general that for self-adjoint $A, B$ their product will be self-adjoint as well, because taking the adjoint is an antiautomorphism - it reverses the order of products:
$$(AB)^* = B^* A^*$$
And for a product of many operators
$$(A_1 A_2 \dots A_n)^* = A_n^*\dots A_2^* A_1^*$$
In particular, this implies that if the product $A_1 A_2 \dots A_n$ is a palindrome ($A_1 = A_n$, $A_2 = A_{n-1}$, etc) and all $A_i$ are self-adjoint, their product will be self-adjoint as well. This is your case precisely, for the product $PTP$ is indeed a palindrome, and $P$ and $T$ are self-adjoint as given.