If the line D: x = 1 is the directrix of the parabola p: y raised to the power of 2 - kx + 8 = 0, then one of the values of k can be:

### Step-by-Step Solution:

### Step 1: Rewrite the parabola equation in standard form

y^2 = kx - 8

### Step 2: Identify vertex form and properties of the parabola

p = \frac{k}{4}

### Step 3: Determine the vertex

\left(\frac{8}{k}, 0\right)

### Step 4: Location of the directrix

The directrix is at:

x = \frac{8}{k} - \frac{k}{4}

Given the directrix is \( x = 1 \):

\frac{8}{k} - \frac{k}{4} = 1

Simplify and solve the equation:

8 - \frac{k^2}{4} = k

\frac{k^2}{4} + k - 8 = 0

k^2 + 4k - 32 = 0

### Step 5: Solve the quadratic equation

Use the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

k = \frac{-4 \pm \sqrt{144}}{2}

k = \frac{-4 \pm 12}{2}

Thus, the solutions are:

k = \frac{-4 + 12}{2} = 4

k = \frac{-4 - 12}{2} = -8

Verify which value matches the directrix condition:

- For \( k = 4 \):

\frac{8}{4} - \frac{4}{4} = 2 - 1 = 1

This matches.

- For \( k = -8 \):

\frac{8}{-8} - \frac{-8}{4} = -1 + 2 = 1

This also matches.

Therefore, \( k = 4 \) satisfies the condition.

\boxed{k = 4}