Given: α and β are distinct roots of \( 2x^2 - 6x + k = 0 \).
From Vieta's formulas, the product of roots is \( \alpha\beta = \frac{k}{2} \) and the sum of roots is \( \alpha + \beta = \frac{-(-6)}{2} = 3 \).
The roots of the equation \( x^2 + px + p = 0 \) are \( r_1 = \alpha + \beta = 3 \) and \( r_2 = \alpha\beta = \frac{k}{2} \).
Using the sum of roots for the second equation: \( r_1 + r_2 = 3 + \frac{k}{2} = -p \) (1).
Using the product of roots for the second equation: \( r_1 \cdot r_2 = 3 \cdot \frac{k}{2} = p \) (2).
Substitute \( p = \frac{3k}{2} \) from (2) into (1): \( 3 + \frac{k}{2} = -\frac{3k}{2} \).
Multiply by 2: \( 6 + k = -3k \Rightarrow 6 = -4k \Rightarrow k = -\frac{3}{2} \).
Substitute \( k = -\frac{3}{2} \) into (2) to find p: \( p = \frac{3}{2} \cdot \left(-\frac{3}{2}\right) = -\frac{9}{4} \).
Calculate \( 8(k - p) \): \( 8\left(-\frac{3}{2} - (-\frac{9}{4})\right) = 8\left(-\frac{3}{2} + \frac{9}{4}\right) \).
Convert to a common denominator: \( 8\left(-\frac{6}{4} + \frac{9}{4}\right) = 8\left(\frac{3}{4}\right) = \boxed{6} \).
Final Answer: 6 (Option C)