This is a good question that I have also had, so I'm answering this even though it is two years old! It looks like your definitions are all fine, but I would use
\begin{equation} \hat{\mathcal{P}}\hat{\psi}^{}{A}\hat{\mathcal{P}}^{-1} = \sum{B} \hat{\psi}{B}^{\dagger}(U{P})_{B, A} \end{equation}
as it simplifies notation. I'll also refer to the particle-hole operator as $\hat{\mathcal{P}}$ to keep my notation clear, but that's just a personal preference.
Using the above definition we implement the transformation of $\hat{\mathcal{P}}$ on $\hat{H}$ as
\begin{equation} \begin{split} \hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} &= \sum_{A, B} \hat{\mathcal{P}} \hat{\psi}{A}^{\dagger} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} H^{}{A, B} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} \hat{\psi}^{}{B} \hat{\mathcal{P}}^{-1}\ & =\sum{A, B} \sum_{C, D} (U_{P}^{})^{\dagger}{A, C} \hat{\psi}{C} H_{A, B} \hat{\psi}{D}^{\dagger} (U{P}^{})^{}{D, B} \ &= \sum{A, B}\sum_{C, D} \delta^{}{C, D}(U{P})^{\dagger}{A, C}H^{}{A, B}(U_{P})^{}{D, B} - \hat{\psi}^{\dagger}{D}(U_{P})^{}{D, B}H^{}{A, B}(U_{P})^{\dagger}{A, C}\hat{\psi}^{}{C} \ & = \sum_{A, B}\sum_{C, D} (U_{P})^{\dagger}{A, C}H^{}{A, B}(U_{P})^{}{C, B} - \hat{\psi}^{\dagger}{D}(U_{P})^{}{D, B}H^{T}{B, A}(U_{P})^{\dagger}{A, C}\hat{\psi}^{}{C} \ & = \sum_{A, B}\left( \delta^{}{A, B}H{A, B} - \sum_{C, D}\hat{\psi}^{\dagger}{D}(U{P})^{}{D, B}H^{T}{B, A}(U_{P})^{\dagger}{A, C}\hat{\psi}^{}{C} \right) \ & = \text{Tr}(H) - \sum_{C, D} \hat{\psi}^{\dagger}{D}H^{}{D, C}\hat{\psi}^{}_{C}, \end{split} \end{equation}
where we have used the anticommutation relation ${\hat{\psi}^{}{A}, \hat{\psi}^{\dagger}{B}} = \delta^{}{A, B}$ between the second and third lines and the unitarity of $U{P}$ between the fourth and fifth lines. We require $\hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} = \hat{H}$ if our Hamiltonian respects PHS, demanding the conditions that
\begin{equation} \begin{split} \text{Tr}(H) &= 0 \ H = -U_{P}&H^{*}U_{P}^{\dagger} \end{split} \end{equation}
i.e. 526
the Hamiltonian is traceless. We have also used the Hermicity of the Hamiltonian in the second equality. Thankfully, these two conditions are complementary, since $H = -U_{P}H^{*}U_{P}^{\dagger}$ implies that our Hamiltonian has a symmetric spectrum. 526