My guess? No.
Here is why. System is adiabatic so only work is being done ON to the system in the compression phase and only work is being done BY the system during the expansion phase. We can calculate the constant pressure work using:
Compression phase: $w= -P_2(V_2-V_1)=-2P_1(V_2-V_1)$
Expansion phase:$w= -P_1(V_3-V_2)$
If we were to assume that the expansion phase does the same magnitude of work as the compression phase did (which it probably doesn't), then at the very least:
$$\lvert w\rvert\ = -P_1(V_3-V_2)=-2P_1(V_2-V_1) \Rightarrow \lvert (V_3-V_2) \rvert\ =2\lvert (V_2-V_1) \rvert\ $$
Because $P_2$ is greater than $P_1$, the system gains more energy during the compression and has to expend less of that energy to fight a smaller external pressure during expansion.
Therefore, I think the last state reached will have $P_1$, $V_3$ (where $V_3>V_1$) , and $T_3$ (where $T_3>T_1$).
Also, since $T_3>T_1$, and since this gas is ideal, we know that not all of the work that was absorbed by the system during phase 1 (compression) was used again as work during expansion. The remainder of that energy was used to increase the internal energy of the system.
The the most important conclusion we can make here is this:
The internal energy of the system before the compression phase is LESS than the internal energy of the system after the expansion phase.