Question about bi-directional pin from Roth reference
I think there should be some other tri-state buffer in the lower signal to prevent this. Is this right?


No. Simple as that; there's no reason not to sense your own voltage. It's not like input circuitry had a low impedance!


  In case of EN=1, what if there is some logic value is put on the bi-directionl pin as an input from external source? 


Two possible things:


External voltage = voltage produced by the buffer: No voltage difference, no current flows, nothing happens
External voltage \$\ne\$ voltage produced by the buffer: you've got your external source trying to work against your internal source. The resulting voltage will be defined by their source impedances (simple voltage divider!). The resulting current can be very high, and potentially damaging to either side 



  This will result in unknown state I think.


No, not really.


  So I think there should be another tri-state buffer on the line between the join and the pin. Is this right?


No. How would that help or even work.