It's a special case of Lucas' theorem (I'm not sure if it was already posted here): If $a_{0}, a_{1}, ... , a_{n}$ and $b_{1}, b_{2}, ... , b_{n}$ are the digits of $a$ and $b$ respectively in $p$-adic representation, then \[\binom ab \equiv \binom{a_{0}}{b_{0}}\binom{a_{1}}{b_{1}}... \binom{a_{n}}{b_{n}}\mod p.\]

Let $ k=pt+u$ with $ 0\le u<p$. Consider the element of $ S_k$ of order $ p$ specified by $ (1,2,3,\ldots,p)(p+1,p+2,\ldots,2p)\ldots((t-1)p+1,(t-1)p+2,\ldots,tp)$, and consider the action of this element on the subsets of $ \{1, \ldots, k\}$ of size $ p$. Clearly all of the orbits are of size 1 or $ p$, and hence the total number of subsets, namely $ C(k,p)$ must be equivalent (modulo $ p$) to the number of fixed points, which is $ t$. Luke See my puzzle blog at http://bozzball.blogspot.com