Let $\displaystyle p>3$ be a prime. $\displaystyle \sum_{i=1}^p\binom {i\cdot p}{p}\cdot\binom {\left(p-i+1\right)p}{p}\equiv ?\mod p^2$.
the answer is $\displaystyle \frac{p(p+1)(p+2)}{6}$ and we don't need the condition $\displaystyle p > 3.$ to prove this let $\displaystyle f(x)=\prod_{k=1}^{p-1}(x-k)=x^{p-1}-a_1x^{p-2} + \cdots -a_{p-2}x + a_{p-1}.$
by the Fermat's little theorem, in $\displaystyle (\mathbb{Z}/p\mathbb{Z})[x],$ we have $\displaystyle f(x)=x^{p-1}-1.$ as a result $\displaystyle (p-1)!=a_{p-1} \equiv -1 \mod p,$ which is Wilson's theorem, and
$\displaystyle a_j \equiv 0 \mod p,$ for all $\displaystyle j \neq p-1.$ thus, for any $\displaystyle n \in \mathbb{N}: \ \binom{np}{p}=\frac{np(np-1) \cdots (np-p+1)}{p!}=\frac{nf(np)}{(p-1)!} \equiv n \mod p^2.$ so your sum, modulo $\displaystyle p^2$,
is equal to $\displaystyle \sum_{i=1}^{p}i(p-i+1)=\frac{p(p+1)(p+2)}{6}. \ \Box$
Remark 1. this method gives another proof for the problem you asked in here.
Remark 2. there's a theorem (i don't remember its name) which says that for any prime number $\displaystyle p > 3: \ a_{p-2} \equiv 0 \mod p^2.$ does anybody know the
name of this theorem? anyway, using this result we get this stronger result that, modulo $\displaystyle p^3,$ your sum for $\displaystyle p > 3$ is still equal to $\displaystyle \frac{p(p+1)(p+2)}{6}.$
ok, i found it! it's Wolstenholme's theorem.