First, find the coefficient of \(x\) in the expansion of \(\left(4x + \frac{10}{x}\right)^3\).

The term that gives \(x\) is \(\binom{3}{2} (4x)^2 \left(\frac{10}{x}\right)^1 = 3 \times 16x^2 \times \frac{10}{x} = 480x\).

Thus, \(p = 480\).

Next, find the coefficient of \(\frac{1}{x}\) in the expansion of \(\left(2x + \frac{k}{x^2}\right)^5\).

The term that gives \(\frac{1}{x}\) is \(\binom{5}{3} (2x)^3 \left(\frac{k}{x^2}\right)^2 = 10 \times 8x^3 \times \frac{k^2}{x^4} = 80k^2 \times \frac{1}{x}\).

Thus, \(q = 80k^2\).

Given \(p = 6q\), we have:

\(480 = 6 \times 80k^2\)

\(480 = 480k^2\)

\(k^2 = 1\)

Therefore, \(k = \pm 1\).