This is an immediate consequence of Lucas' theorem. We give an elementary proof as follows.

Let $n = ap + r$ where $0 \le r \le p - 1$. Then we have
$$\binom{n}{p} = \frac{n(n-1)\cdots(n-p+1)}{p!} \\ = \frac{(ap+r)(ap+r-1)\cdots(ap+1)(a)(ap-1)(ap-2)\cdots(ap+r-p+1)}{(p-1)!}$$
Note that the numerator is congruent to
$$r(r-1)\cdots(1)(a)(p-1)(p-2)\cdots(r+1) = a \cdot (p-1)!$$
modulo $p$. Therefore, we find that
$$\binom{n}{p} \equiv a = \left\lfloor \frac{n}{p} \right\rfloor \pmod{p}.$$
This proves the desired result.