what would be p-v diagram for irreversible isothermal expansion
If you have constant external pressure $P_{ext}$, the work on the PV diagram will be minus the area of a rectangle that is $P_{ext}$ high and $V_{final} - V_{initial}$ long. 



When you want to show work as an area, you must plot external pressure on the y-axis of a PV diagram, not gas pressure. In the special case of a reversible expansion or compression, the gas pressure $P$ is equal to the external pressure $P_{ext}$ at every point during the process, so you are still plotting external pressure if you're relating work to the area under the curve. 

Why does the external pressure matter in the calculation of expansion work in irreversible processes, and not the gas pressure? 

The reasoning goes like this:

expansion work = opposing force x distance moved

opposing force = external pressure x area

volume change = area x distance moved

so

expansion work  = -(external pressure) x (volume change) 

The minus sign comes from our decision to make work negative when energy leaves the system in an expansion, and positive when energy enters the system from the surroundings in a compression. This is the convention used in chemistry; you'll find the opposite convention used in physics!