The ${\mathrm{P}}_z$ orbital corresponds to the magnetic quantum m equal to

$\begin{array}{l}{\mathbf{p}}_z:m_l=0\\ {\mathbf{p}}_{\mathrm{x}}:50\mathrm{\%}{\mathrm{m}}_{\mathrm{l}}=+1\text{ and }50\mathrm{\%}{\mathrm{m}}_{\mathrm{l}}=-1\\ {\mathbf{p}}_{\mathrm{y}}:50\mathrm{\%}{\mathrm{m}}_{\mathrm{l}}=+1\text{ and }50\mathrm{\%}{\mathrm{m}}_{\mathrm{l}}=-1\end{array}$

This occurs because the ${\text{p}}_x$ and ${\text{p}}_y$ orbitals are a linear combination of the ${\text{p}}_{+1}$ and${\text{p}}_{-1}$ wave functions, not the eigenstates of the ${\hat{L}}_z$ operator.