I agree: It's not an isothermal process. It merely starts and ends at the same temperature.
It consists of two steps: an isochoric (constant-V) process, in which the pressure is decreased by cooling the gas at constant V, followed by an isobaric process, in which the volume is increased by heating the gas at constant p, returning it to its original temperature (but not its original pressure or volume).
As to what Chet Miller writes:
In short, the answer given by @theorist isn't even close to being correct.
"Isn't even close to being correct" is actually a good description of his own comment.
I stand by what I said. The OP posted a graph that showed how the pressure and volume of an ideal gas varied during a process. I can only answer based on what was posted. Logically, for the first leg, if p is going down at constant V, the system must be cooling. And for the second leg, if V is going up at constant p, the system must be heating. [We're assuming the system is closed.] It is easy to accomplish this: Simply put the system in contact with a lower-T bath while keeping V constant, and then put the system in contact with a bath at the original T, keeping p constant.
For instance, we could start at 300 K, 1 L, and 2 atm. Then we could gradually cool at constant V to 150 K, 1 L, 1 atm. Then we could gradually heat at constant p to 300 K, 2 L, 1 atm.
As for his niggle that T and p can never be defined during an irreversible process, that's an unfortunate and absolutist position. In practice, it certainly is possible to define and measure T and p during irreversible processes, particularly if we make the change slow enough that the system can relax close enough to an equilibrium state such that our measuring devices can't tell the difference—or that the difference is less than the precision with which we care to report our measurements. [This is not to be confused with a true reversible process, which is an idealization that can't exist is the real world.]
And the diagram, by showing the pressure of the system throughout the process, is telling us just that: That the system is sufficiently relaxing during the process to have a measurable p. I.e., yes, normally the default assumption during an irreversible process is that the intensive variables are undefined (this is sometimes indicated using a dashed line). But here the diagram, by using an unbroken solid line to describe p throughout the process, is telling us p is measurable.
Consider the converse: If we hold to such an absolutist view, that systems must be at equilibrium for us to be able to describe their temperature and pressure, then we can never measure or discuss the temperature or pressure of any real-world system, since no real-world system is ever truly at equilibrium. So essentially all the temperature and pressure measurements in all the world's scientific literature are invalid.
Note that this would also prevent us from talking about the temperature or pressure change in a Joule-Thomson expansion, since such an expansion is a steady-state process so it is, by design, never at equilibrium.
I see from Chet's profile that he's a retired chemical engineer. To be logically consistent with the view he presents here, anytime his colleagues or bosses asked him about the temperature or pressure of a system, he'd always have to say: "No real-world system is ever at equilibrium, so they have no definable temperature or pressure." Do you think that's what he always said?
The process Chet is ascribing to the question is different from what's pictured. He's taking the diagram to represent the V of the system, but the p of the surroundings. Based on that assumption, his description is correct. But, while that may be the intent of the question, there's nothing in the presented diagram to indicate that's the case. He is, as the lawyers say, "assuming facts not in evidence."
As to his other comment:
Really???? So, in stage 1 you start out with two entries, the gas and the reservoir, at the same temperaturer. Then, spontaneously, heat flows from the gas (at constant volume) to the reservoir so that, at the end of stage 1, they are at different temperatures (with the gas colder). Do you really think that this is consistent with the 2nd law of thermodynamics?
I applaud his restraint in limiting himself to four question marks. Had he used five, I would be concerned.
He's attributing to me a bizzare process that bears no relationship to anything I said, nor to what I had in mind, nor to what's pictured by the diagram. I have no idea how or why he came up with this. It's so strange it took me a while to figure out what he was trying to describe, and almost seems like a straw man. To adopt Chet's parlance: Really??? [I'm not as well-trained as Chet in hyperbole, so I can only go to three question marks, max.]