The general formula for the efficiency of any cycle is
$$η=\frac {W_{net}}{Q_{in}}$$
Where $W_{net}=Q_{in}-Q_{out}$
That includes your cycle as well as the Carnot Cycle.
But the Carnot Cycle consists of two reversible isothermal (constant temperature) processes and two reversible adiabatic (constant entropy) processes. For the two isothermal reversible processes
$$Q_{in}=T_{B}\Delta S$$
$$Q_{out}=T_{A}\Delta S$$
Where $T_A$ is a low temperature thermal reservoir where heat is rejected, $T_B$ is a high temperature thermal reservoir where heat $Q_{in}$ is added, and the magnitude of the change in entropy, $\Delta S$, is the same for both isothermal processes.
Plugging the last two equations into the general formula for efficiency gives you the Carnot Cycle efficiency of
$$η=1-\frac{T_A}{T_B}$$
Which is the greatest efficiency possible for a heat engine operating between two temperatures.
In my case Can I considerate that to make AB transformation my gas is
in contact with a source at temperature 𝑇𝐴. If I can where is the
problem? If I'm agree so I can use 1−𝑇𝐴/𝑇𝐵
No you can't. The problem is temperature is not constant during process AB and the Carnot efficiency only applies to an isothermal (constant temperature) processes. For process AB the gas has to be in contact multiple temperature sources each infinitesimally higher than the previous in going from A to B.
For an ideal gas we have
$$\frac{P_{A}V_{A}}{T_{A}}=\frac{P_{B}V_{B}}{T_{B}}$$
Since it is a constant volume process, $V_{A}=V_{B}$, therefore
$$\frac{P_A}{T_A}=\frac{P_B}{T_B}$$
Since $P_{B}>P_{A}$, we have $T_{B}>T_{A}$
The temperature for process AB is only $T_A$ at point A. To apply the Carnot cycle efficiency heat has to be added at the same constant temperature throughout the process.
Hope this helps.