Mongolian Mathematical Olympiad

The sum $\sum_{k=0}^{p} (-1)^k \binom{p}{k} \binom{p+k}{k}$ is the coefficient of $x^p$ in the expansion of
$$\sum_k \binom{p}{k} (x-1)^{p+k}.$$
This can be rewritten as
$$\left\{ \sum_{k=0}^{p} \binom{p}{k} (x-1)^j \right\} (x-1)^p = x^p (x-1)^p$$
Hence the sum is actually $(-1)^p = -1$. Hence
$$\sum_{k=0}^{p} (-1)^k \binom{p}{k} \binom{p+k}{k} \equiv -1 \pmod{p^3}$$
clearly holds.