To find the coefficient of \(x^3\) in the expansion of \((1 - px)^5\), we use the binomial theorem:

\((1 - px)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-px)^k\).

The term containing \(x^3\) is when \(k = 3\):

\(\binom{5}{3} (1)^{5-3} (-px)^3 = \binom{5}{3} (-1)^3 p^3 x^3\).

\(\binom{5}{3} = 10\), so the coefficient of \(x^3\) is:

\((-1) \cdot 10 \cdot p^3 = -10p^3\).

We are given that this coefficient is \(-2160\):

\(-10p^3 = -2160\).

Dividing both sides by \(-10\):

\(p^3 = 216\).

Taking the cube root of both sides:

\(p = 6\).