A

Step 1: Understand the process of the cycle. We have three main processes:
1. Adiabatically expanding from (p₁, V₁) to (p₂, V₂)
2. Isobarically compressing from (p₂, V₂) to (p₂, V₁)
3. Raising pressure at constant volume from (p₂, V₁) back to (p₁, V₁)

Step 2: Use the First Law of Thermodynamics. For an adiabatic process, we have:
$$ Q = 0 = riangle U + W $$
Where $W$ is the work done by the gas and $ riangle U$ is the change in internal energy.

Step 3: For an adiabatic process, using the relation:
$$ p V^{ ext{γ}} = ext{constant} $$
We can relate the states during the adiabatic expansion:
$$ rac{p_1 V_1^{ ext{γ}}}{p_2 V_2^{ ext{γ}}} = 1 ightarrow p_1 V_1^{ ext{γ}} = p_2 V_2^{ ext{γ}} $$
From this, we derive:
$$ V_2 = V_1 igg( rac{p_1}{p_2} igg)^{rac{1}{ ext{γ}}} $$

Step 4: Calculate the work done during the adiabatic process. For an ideal gas:
$$ W_{ad} = rac{p_1 V_1 - p_2 V_2}{ ext{γ} - 1} $$
Substituting $V_2$ we get:
$$ W_{ad} = rac{p_1 V_1 - p_2 igg( V_1 igg( rac{p_1}{p_2} igg)^{rac{1}{ ext{γ}}} igg)}{ ext{γ} - 1} $$

Step 5: The heat added during the isobaric process is:
$$ Q = p_2 imes (V_1 - V_2) $$
And work done in isobaric compression is:
$$ W_{iso} = p_2 (V_1 - V_2) $$
For the third step (constant volume): no work is done as $W = 0$. The heat is provided to change the pressure.

Step 6: The efficiency is then defined as:
$$ ext{η} = rac{Q_{out} - Q_{in}}{Q_{out}} $$
By plugging in the values from the process above, after manipulation, we arrive at:
$$ ext{η} = 1 - rac{ ext{γ} (V_2/V_1 - 1)}{(p_1/p_2 - 1)} $$

Conclusion: Therefore, the efficiency of the cycle is given by the expression provided in the problem statement.