Equivalent resistance in BJT small-signal : Rpi in parallel with Beta*Ib
TThe dynamic resistance \$r_e\$ is nothing else than the resistance of the parallel combination of \$r_{\pi}\$ and the current source (nomenclature as in the figure):
Test voltage: \$v_T=i_T r_e\$ with \$i_T=i_b + \beta i_b\$ and with \$i_b=\frac{v_T}{r_{\pi}}\$.
From this you can solve for \$r_e=\frac{r_{\pi}}{\beta+1}\$
By the way (physical interpretation): the resistance \$r_e\$ is nothing else than the inverse transconductance \$g_e\$ for the transistor: \$r_e=\frac{1}{g_e}=\frac{dV_{be}}{dI_e}\$.
From the diagram you can derive that \$r_e\$ is in parallel with the sum of \$R_E\$ and \$R_s||R_L\$