Let $p$ be a prime number.
Let $k$ and $m$ be positive integers such that $k \ge m$ and $p-1$ does not divide $k-m$.
It seems to be true that $$ (-1)^{m-\lfloor \frac{k-m}{p-1}\rfloor}\sum_{i\ge 1}(-1)^i{{k-1+i(p-1)}\choose{k-1}}{m\choose{\lfloor \frac{k-m}{p-1}\rfloor+i}} \equiv 0 \pmod p.$$ and for any prime $p$ and any natural numbers $q, n$ such that $p$ does not divide $n$, it holds that $$\sum_{i\ge q+1}(-1)^i {\lfloor \frac{n}{p}\rfloor p \choose i}{n-1+i(p-1) \choose n-1+q(p-1)} \equiv 0\pmod p. $$
$\lfloor \frac{k-m}{p-1}\rfloor$ stands for the largest integer smaller than or equal to $\frac{k-m}{p-1}$.
I have made numerical verification for various $k,m,p$.
For $p=3$ and $k-m=2q+1$ this is : $$ \sum_{i\ge q+1}(-1)^{m-i}{{m+2i}\choose{m+2q}}{m\choose i} \equiv 0 \pmod 3$$ which looks more easy, but still I did not succeed. Edit: I know ho to make it for the case $q=0$, where the lhs is 3 times the Jacobsthal numbers $a_m$, such that $a_m=a_{m-1}+2a_{m-2}$ with $a_0=a_{1}=1$. But in the next case, $q=1$, the lhs produces $3 \cdot \{5,21,84,265,786,...\}$. The sequence of numbers (conjecturally integers) in the braces looks quite mysterious.
Any ideas?
What I have tried: backward induction on $q=\lfloor \frac{k-m}{p-1}\rfloor$ for fixed $m$ and fixed $r$ (residue class of $k-m$ modulo $p-1$). I had this induction idea because it is trivially true when $q \ge m$ and it easy to see that this is true also for $q=m-1$. In that case, the lhs is ${{(m-1)p+r}\choose{p-1}}$ which is ${{r}\choose{p-1}}$ modulo $p$, which is zero since $r<p-1$.
Edit (03/29/20): I have found a (lengthy) proof. It makes use of Stirling numbers.