I understand that if a Hamiltonian remains invariant under the following transformations then it is PT invariant,
\begin{eqnarray} \mathrm{Parity \; reversal:} \; \; \hat{p} \to -\hat{p} \; \; \mathrm{ and } \; \; \hat{x} \to -\hat{x} \\ \mathrm{Time \; reversal:} \; \; \hat{p} \to -\hat{p}, \; \hat{x} \to \hat{x}, \; \; \mathrm{ and } \; \; i \to -i. \end{eqnarray}
However, if during parity reversal we change the sign on momentum, and during time reversal, we change it back, what's the use of performing these rotations on $\hat{p}$? Similarly, the momentum operator contains a first space derivative. Does changing $\hat{p} \to -\hat{p}$ imply that we take $\hat{x} \to -\hat{x}$ (or does the effect cancel out when both $\hat{p}$ and $\hat{x}$ are reversed)?