To find the coefficient of \(x^3\) in the expansion of \(\left(p + \frac{1}{p}x\right)^4\), we use the binomial theorem:

\(\binom{4}{3} \cdot p^{4-3} \cdot \left(\frac{1}{p}\right)^3 \cdot x^3 = 4 \cdot p \cdot \frac{1}{p^3} \cdot x^3 = \frac{4}{p^2}x^3\).

The coefficient of \(x^3\) is \(\frac{4}{p^2}\).

We are given that this coefficient is 144:

\(\frac{4}{p^2} = 144\).

Solving for \(p^2\), we get:

\(p^2 = \frac{4}{144} = \frac{1}{36}\).

Taking the square root of both sides, we find:

\(p = \pm \frac{1}{6}\).