What will be pressure of gas measured by manometer if mercury starts to rise on the other side? What will be pressure of gas measured by manometer if mercury starts to rise on the other side?

The $P_a$ side is open to the gas container and the $P_0$ side is open to the atmosphere with atmospheric pressure $P_0$.
The pressure of gas $P$ here is $P_0 + \rho g H$ where $H$ is as given in image and other symbols have usual meanings.
Now here is my question,
What would be the pressure of a gas if the pressure of gas was high enough to cause the liquid (mercury here) to move down the U shaped tube and rise up on the other side?
 A: If the gas pressure $P_a$ increases from the diagram, the mercury level on the right will fall and that on the left will rise. This will increase the value of $H$ so the pressure of the mercury pushing back on the gas also increases, untill $P_0+\rho gH=P_a$ again.
If the gas pressure $P_a$ is strong enough to push the mercury level to the bottom of the manometer tube, then $H$ has reached its maximum possible value $H_{max}$, which is the length of mercury in the manometer (straight plus curved sections). Even if the mercury rises further up on the left the difference $H$ between the 2 levels of mercury cannot increase any more. So the pressure pushing back on the gas cannot increase any more. 
If $P_0+\rho gH_{max} \lt P_a$ then the mercury will be pushed further and further up on the left until all of it is forced out of the top of the manometer tube, however high that is. 
In short, if the gas can push the mercury down to the very bottom of the bend in the manometer, then it will continue pushing the mercury all the way out of the manometer.
