## Appendix A  
**Proof of Proposition 1. ** The terms \(B_{1}\) and \(B_{2}\) can be calculated with the help of (2) and (3). The former becomes  
\[ \begin{split}& B_{1}=\Pr\left(\gamma_{SD1,k*}^{PDMA}<\varepsilon_{ 1},\gamma_{SD1\leftarrow 2,k*}^{PDMA}<\varepsilon_{1}\right)\\ &=\underbrace{\left(1-\Pr\left(\gamma_{SD1,k*}^{PDMA}\geq \varepsilon_{1}\right)\right)}_{B_{11}}\underbrace{\left(1-\Pr\left(\gamma_{ SD1\leftarrow 2,k*}^{PDMA}\geq\varepsilon_{1}\right)\right)}_{B_{12}}\.\end{split}  \tag{10}\]  
Therefore, \(B_{11}\) can be expressed as  
\[ \begin{split}& B_{11}=1-\Pr\left(\gamma_{SD1,k}^{PDMA}\geq \varepsilon_{1}\right)\\ &=1-\Pr\left(\left|g_{1,k}\right|^{2}\geq\frac{\varepsilon_{1} \omega\rho|f_{1}|^{2}+\varepsilon_{1}}{\varepsilon_{1}\rho-\varepsilon_{1} \omega_{2}\rho}\right)\\ &=1-\frac{1}{\lambda_{f_{1}}}\int_{0}^{\infty}\sum\limits_{k=1}^{ K}\left(\begin{array}{c}K\\ k\end{array}\right)(-1)^{k-1}\exp\left(-\frac{(\varepsilon_{1}\omega\rho x+ \varepsilon_{1})k}{(\varepsilon_{1}\rho-\varepsilon_{1}\omega_{2}\rho) \lambda_{1}}-\frac{x}{\lambda_{f_{1}}}\right)dx\\ &=1-\sum\limits_{k=1}^{K}\left(\begin{array}{c}K\\ k\end{array}\right)(-1)^{k-1}\vartheta_{1},\end{split}  \tag{11}\]  
where \(\vartheta_{1}=\frac{(\alpha_{1}\rho-\varepsilon_{1}\omega_{2}\rho)\lambda_{1}} {\varepsilon_{1}\omega\rho\lambda_{f_{1}}+(\alpha_{1}\rho-\varepsilon_{1} \omega_{2}\rho)\lambda_{1}}\exp\left(-\frac{\varepsilon_{1}k}{(\alpha_{1} \rho-\varepsilon_{1}\omega_{2}\rho)\lambda_{1}}\right)\). It is noted that \(\lambda_{1}\) is average channel gain of \(g_{1,k}\), then \(\lambda_{2}\) is average channel gain of \(g_{2,k}\), \(\lambda_{f1},\lambda_{f2}\) are average channel gain of \(f_{1},f_{2}\), respectively; \(\lambda_{h1},\lambda_{h2}\) are average channel gains of \(h_{1},h_{2}\), respectively. Similarly, \(B_{12}\) can be expressed by  
\[ \begin{split}& B_{12}=1-\Pr\left(\gamma_{SD1\gets 2,k}^{PDMA} \geq\varepsilon_{1}\right)\\ &=1-\Pr\left(\left|g_{2,k}\right|^{2}\geq\frac{\varepsilon_{1} \omega\rho|f_{2}|^{2}+\varepsilon_{1}}{\varepsilon_{1}\rho-\varepsilon_{1} \omega_{2}\rho}\right)\\ &=1-\frac{1}{\lambda_{f_{2}}}\int_{0}^{\infty}\sum\limits_{k=1}^{ K}\left(\begin{array}{c}K\\ k\end{array}\right)(-1)^{k-1}\exp\left(-\frac{(\varepsilon_{1}\omega\rho y+ \varepsilon_{1})k}{(\varepsilon_{1}\rho-\varepsilon_{1}\omega_{2}\rho) \lambda_{2}}-\frac{y}{\lambda_{f_{2}}}\right)dy\\ &=1-\sum\limits_{k=1}^{K}\left(\begin{array}{c}K\\ k\end{array}\right)(-1)^{k-1}\vartheta_{2},\end{split}  \tag{12}\]  
where \(\vartheta_{2}=\frac{(\alpha_{1}\rho-\varepsilon_{1}\omega_{2}\rho)\lambda_{2}} {(\alpha_{1}\rho-\varepsilon_{1}\omega_{2}\rho)\lambda_{2}+\varepsilon_{1} \omega\rho\lambda_{f_{2}}}\exp\left(-\frac{\varepsilon_{1}k}{(\alpha_{1}\rho- \varepsilon_{1}\omega_{2}\rho)\lambda_{2}}\right)\). From (A2) and (A3) we find the expression \(B_{1}\) as  
\[B_{1}=\left(1-\sum\limits_{k=1}^{K}\left( \begin{array}{c}K\\ k\end{array} \right)(-1)^{k-1}\theta_{1}\right)\left(1-\sum\limits_{k=1}^{K} \left( \begin{array}{c}K\\ k\end{array} \right)(-1)^{k-1}\theta_{2}\right).\] (A4)  
In similar way, \(B_{2}\) can be computed as  
\[ \begin{array}{l}B_{2}=\Pr\left(\max\left\{\gamma_{SD1,k*}^{PDMA}\chi_{D1}^{ PDMA}\right\}<\varepsilon_{1},\gamma_{SD1\gets 2,k*}^{PDMA}>\varepsilon_{1}\right)\\ =\Pr\left(\gamma_{SD1,k*}^{PDMA}<\varepsilon_{1}\right)\Pr\left(\chi_{D1}^{ PDMA}<\varepsilon_{1}\right)\Pr\left(\gamma_{SD1\gets 2,k*}^{PDMA}>\varepsilon_{1}\right)\\ =\underbrace{\left(1-\Pr\left(\gamma_{SD1,k*}^{PDMA}\geq \varepsilon_{1}\right)\right)}_{B_{21}}\underbrace{\left(1-\Pr\left(\chi_{D1 }^{PDMA}\geq\varepsilon_{1}\right)\right)}_{B_{22}}\\ \times\underbrace{\Pr\left(\gamma_{SD1\gets 2,k*}^{PDMA}> \varepsilon_{1}\right)}_{B_{23}}.\end{array} \] (A5)  
In this step, \(B_{21}\) is formulated as  
\[