Solution:

Answer: $p > 1$ and $p < -3$

First, we note that
$$x^{3} + p x^{2} - p x - 1 = (x - 1)\left(x^{2} + (p + 1)x + 1\right)$$
Hence, $x^{2} + (p + 1)x + 1$ has two distinct roots. Consequently, the discriminant of this equation must be positive, so $(p + 1)^{2} - 4 > 0$, so either $p > 1$ or $p < -3$.

However, the problem specifies that the quadratic must have distinct roots (since the original cubic has distinct roots), so to finish, we need to check that $1$ is not a double root—we will do this by checking that $1$ is not a root of $x^{2} + (p + 1)x + 1$ for any value $p$ in our range. But this is clear, since $1 + (p + 1) + 1 = 0 \Rightarrow p = -3$, which is not in the aforementioned range. Thus, our answer is all $p$ satisfying $p > 1$ or $p < -3$.