As $ p $ is a prime number, it follows that the modular inverses of $ 1,2, \ldots, p-1 $ all exist. We claim that $ n^{-1} \cdot (n+1)^{-1} \equiv n^{-1} - (n+1)^{-1} \pmod{p} $ for $ n \in \{1,2, \ldots, p-2\} $, in analogue with the formula $ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} $. Indeed, multiplying both sides of the congruence by $ n(n+1) $, we find that $$1 \equiv n(n+1) \cdot (n^{-1} - (n+1)^{-1}) \equiv (n+1) - n \equiv 1 \pmod{p},$$as desired. Thus, \begin{align*}&1^{-1} \cdot 2^{-1} + 2^{-1} \cdot 3^{-1} + 3^{-1} \cdot 4^{-1} + \cdots + (p-2)^{-1} \cdot (p-1)^{-1} \\ &\equiv 1^{-1} - 2^{-1} + 2^{-1} - 3^{-1} + \cdots - (p-1)^{-1} \pmod{p}.\end{align*}This is a telescoping series, which sums to $ 1^{-1} - (p-1)^{-1} \equiv 1 - (-1)^{-1} \equiv \boxed{2} \pmod{p} $, since the modular inverse of $ -1 $ is itself.