Here in this circuit the circuit generates a quasi-stable state transition from the stable state.Say the output of the V0= L+ I mean HIGH or Vsat.Now feedback fraction=b=R1/(R1+R2).Now as the output voltage is L+ then voltage at the C terminal is bL+.And as the diode D1 is getting High voltage at the P side so diode D1 is ON.So the voltage at the terminal B is just one dide drop say VD1.So the output remains at L+(I mean HIGH).So thisnis the stable state.BUT diode D2 is OFF as R4 is high.So this is the stable state of the monostable multivibrator.
How we go from stable state to quasi-stable state?
We apply a pulse at the trigger terminal. So then diode D2 will now conduct and the voltage at the C terminal will drop. Now when the voltage at the C terminal drops below voltage at B terminal then the output changes to L- or LOW state so this way HIGH to LOW(quasi-stable) state transition occur.
When back to stable state again from quasi-stable state.
Answer is when the output is at Low voltage (L-) then the voltage at terminal C is bL-.Now as the trigger is removed (because trigger is a very short duration pulse).So now as the diode D1 p-sde has low voltage so DIO D1 will be OFF and the capacitor C1 will try to charge towards L-.So a time will come when the voltage at the B terminal will be less than bL- and then output of the OPAMP go to L+ state again. But still this is not the stable state. Stable state will be achieved when the diode D1 will be On again.
What is recovery time
When the capacitor C1 will discharged as output (point A ) is L+. So C1 will discharge first and then when voltage at point B is more than diode potential barrier then diode D1 will be ON and then current will be flown via A to D1 and voltage at point B is again one diode drop.So stable state achieved. The time required to discharge the C1 is called recovery time.