\(The first term of the geometric progression is a = p and the common ratio is r = 2.\)

The sum of the first n terms of a geometric progression is given by:

\(S_n = \frac{a(r^n - 1)}{r - 1}\)

Substituting the values, we have:

\(S_n = \frac{p(2^n - 1)}{2 - 1} = p(2^n - 1)\)

We are given that:

\(p(2^n - 1) > 1000p\)

Dividing both sides by p (since p is positive), we get:

\(2^n - 1 > 1000\)

Adding 1 to both sides gives:

\(2^n > 1001\)