The given problem involves the dissociation of NO gas and the determination of the value of Kp for the reaction 2NO ⟶ N2 + O2. Let's analyze the problem step by step.

**Given Information:**
- Volume of the tube = 0.25 L
- Moles of NO dissociated = 4
- Degree of dissociation (α) = 10%

**Step 1: Calculate the moles of NO dissociated:**
The degree of dissociation (α) is given as 10%, which means that only 10% of the initial moles of NO will dissociate. Therefore, the moles of NO dissociated can be calculated as:
Moles dissociated = Moles of NO × α
= 4 × 0.10
= 0.4 moles

**Step 2: Calculate the moles of NO, N2, and O2 at equilibrium:**
Since two moles of NO react to form one mole each of N2 and O2, the moles of N2 and O2 formed at equilibrium will also be equal to the moles of NO dissociated.
Moles of N2 = Moles of O2 = 0.4 moles
Moles of NO remaining = Moles of NO - Moles dissociated
= 4 - 0.4
= 3.6 moles

**Step 3: Calculate the partial pressures of NO, N2, and O2 at equilibrium:**
Partial pressure (P) can be calculated using the ideal gas law equation:
P = (n/V) × (R × T)
where n is the number of moles, V is the volume, R is the ideal gas constant, and T is the temperature.

For NO at equilibrium:
P(NO) = (moles of NO remaining / volume of the tube) × (R × T)
= (3.6 moles / 0.25 L) × (R × T)
= 14.4 × (R × T)

For N2 and O2 at equilibrium:
P(N2) = P(O2) = (moles of N2 or O2 / volume of the tube) × (R × T)
= (0.4 moles / 0.25 L) × (R × T)
= 1.6 × (R × T)

**Step 4: Write the expression for Kp:**
The expression for Kp for the reaction 2NO ⟶ N2 + O2 is given by:
Kp = (P(N2) × P(O2)) / (P(NO)^2)

**Step 5: Substitute the calculated values into the Kp expression:**
Kp = (1.6 × (R × T) × 1.6 × (R × T)) / (14.4 × (R × T))^2
= (2.56 × (R × T)^2) / (207.36 × (R × T)^2)
= 2.56 / 207.36

**Step 6: Simplify the expression:**
Kp = 0.0124

Therefore, the value of Kp for the reaction 2NO ⟶ N2 + O2, given a degree of dissociation of 10%, is 0.0124.