Since perfect complexes are dualizable, for every perfect complex $P$ and any complex $Q$ we have $$\mathrm{hom}(P,Q)\cong \mathrm{hom}(P,1)\otimes Q\,.$$ Moreover $\mathrm{hom}(P,1)$ is perfect too (since for qcqs schemes perfect=dualizable). In particular, by taking $Q=\omega$ the dualizing sheaf this shows that the image of the duality $D(-):=\mathrm{hom}(-,\omega)$ is exactly $\mathrm{Perf}(X)\otimes\omega$.

Since perfect complexes are dualizable, for every perfect complex $P$ and any complex $Q$ we have $$\mathrm{hom}(P,Q)\cong \mathrm{hom}(P,1)\otimes Q\,.$$ Moreover $\mathrm{hom}(P,1)$ is perfect too (since for qcqs schemes perfect=dualizable). In particular, by taking $Q=\omega$ the dualizing sheaf this shows that the image of the duality $D(-):=\mathrm{hom}(-,\omega)$ is exactly $\mathrm{Perf}(X)\otimes\omega$.

I don't think you can imply any more than this (in particular it's unclear to me why you would expect this to give an equivalence $\mathrm{Perf}(X)^{op}\cong \mathrm{Perf}(X)$ different from the standard duality.