The coefficient of \(x^2\) in \(\left( 1 + \frac{2}{p} x \right)^5\) is given by:

\(\binom{5}{2} \left( \frac{2}{p} \right)^2 = 10 \times \frac{4}{p^2} = \frac{40}{p^2}\)

The coefficient of \(x^2\) in \((1 + px)^6\) is given by:

\(\binom{6}{2} (p)^2 = 15p^2\)

Setting the sum of these coefficients equal to 70:

\(\frac{40}{p^2} + 15p^2 = 70\)

Multiply through by \(p^2\) to clear the fraction:

\(40 + 15p^4 = 70p^2\)

Rearrange to form a quadratic in \(p^2\):

\(15p^4 - 70p^2 + 40 = 0\)

Let \(q = p^2\), then:

\(15q^2 - 70q + 40 = 0\)

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

\(q = \frac{70 \pm \sqrt{70^2 - 4 \times 15 \times 40}}{30}\)

\(q = \frac{70 \pm \sqrt{4900 - 2400}}{30}\)

\(q = \frac{70 \pm \sqrt{2500}}{30}\)

\(q = \frac{70 \pm 50}{30}\)

This gives \(q = 4\) or \(q = \frac{2}{3}\).

Thus, \(p^2 = 4\) or \(p^2 = \frac{2}{3}\).

Therefore, \(p = \pm 2\) or \(p = \pm \frac{\sqrt{2}}{3}\).