Common collector - bias resistor?
I think the purpose of this \$130\Omega\$ resistor is to set a rough threshold for the turn on point of the bipolar transistor.
Considering the original buffer:

A vary small base current suffices already to turn on the bipolar, and if it is large enough it will drive the NPN into saturartion, meaning:
$$V_E = V_1 - V_{CE}$$
Then, the base voltage can be calculated
$$V_B = V_E + V_{BE}$$
Considering a very low \$V_{CE}\$ (deep saturation), the \$V_B\approx V_1 + V_{BE}\$.
Now if you add a base-emitter resistor, things change a bit. You now have a resistor bypassing the base-emitter diode, meaning that as long its voltage is smaller than \$V_{BE}\$ of the transistor very little current will flow through it.
As you increase the base current the voltage drop across the resistor \$130\Omega\$ starts to increase, and if it is high enough it will turn on the transistor and drive it into saturation. So basically it is there to set a threshold current for turning it on. The \$130\Omega\$ probably was chosen according to the following:
$$R_{BE} = \dfrac{V_{BE}}{I_{B,TH}}   \approx \dfrac{0.7V}{5.4mA}=129\Omega$$

Here is a small simulation of the above circuit, where the x-axis represents the current source driving the transistor: