You should define your notations. But let me make a guess: $e_T(-\infty,t)$ is the unitary projector over the subspace $H(T,t)$, invariant under $T$, on which $T\le tI$. And an inequality between projectors means that the range of the first one is included in that of the second. Right ?

Is yes, then the answer to your question is No. Because there is no reason why $T$ and $pTp$ have an invariant subspace in common. Take the finite dimensional case in dimension $2$, and $$T=\begin{pmatrix} -1 & a \\\\ a & 1 \end{pmatrix},$$ with $a\ne0$. Take $p$ the projection onto the first axis. Finally, take $t=-\frac12$. Then $e_{pTp}(-\infty,t)=p$, while $e_T(-\infty,t)$ is the projection onto a line which is not an axis (the eigenvalues of $T$ are $\pm\sqrt{1+a^2}$). Thus both projectors are not comparable.

You should define your notations. But let me make a guess: $e_T(-\infty,t)$ is the unitary projector over the subspace $H(T,t)$, invariant under $T$, on which $T\le tI$. And an inequality between projectors means that the range of the first one is included in that of the second. Right ?

If yes to that, then the answer to your question is No. Because there is no reason why $T$ and $pTp$ have an invariant subspace in common. Take the finite dimensional case in dimension $2$, and $$T=\begin{pmatrix} -1 & a \\\\ a & 1 \end{pmatrix},$$ with $a\ne0$. Take $p$ the projection onto the first axis. Finally, take $t=-\frac12$. Then $e_{pTp}(-\infty,t)=p$, while $e_T(-\infty,t)$ is the projection onto a line which is not an axis (the eigenvalues of $T$ are $\pm\sqrt{1+a^2}$). Thus both projectors are not comparable.