AC base resistance of BJT
I can understand your confusion. The problem is as follows:
The resistive symbol r´e in the small-signal equivalent diagram is, in reality, no resistor at all. It symbolizes the transconductance gm=d(Ic)/d(Vbe)=1r´e  which constitutes the connection between the control input (base-emitter pn junction) and the output (current Ic). The transconductance is the slope of the Ic=f(Vbe) curve - determined at the DC quiescent current.
During calculation we can assume that r´e=1/gm is a small-signal (dynamic) resistance - and that is the reason that there is such an equivalent diagram. But this has no physical meaning - it does not reflect the physical working principle of the transistor.
As far as r´b is concerned, it is a kind of differential &quot;path resistance&quot; between the outer base node and an &quot;inner point&quot; of the base region - and it can be neglected in practice (in comparison with the much larger base-emitter resistance rbe=r_pi)
Personal comment: That is the reason I do not like at all the so-called T-model. It can (as we have seen here) create confusion.
Example: Looking at the most right model, it seems that the input resistance at the base would be r´e=1/gm (in practice some tenth of volts). However, that is completely wrong - normally, some kOhms (because the contribution of the generated current is to be considered).