I have ran some experiments on Magma and I believe this is true: if $p$ is a prime number and $k, a \geq 0$ are integers then
$${kp \choose ap} \equiv {k \choose a} \mod p^2.$$
Can anyone think of how to prove this? A combinatorial proof is given in a comment here, but I am curious if there is an algebraic proof.
I know how to prove congruence mod $p$: it follows by using the binomial theorem on the identity $$(1+x)^{kp} = (1+ x^p)^k$$ in $\mathbb{F}_p[x]$, which in turn is true by Frobenius.
Ok here it is (for $p^3$, an even stronger case):
For this, I will use this nice lemma:
For $n\in\mathbb{N}$ and $p\in\mathbb{P}$, $p>3$: $$\binom{np-1}{p-1}\equiv 1\pmod{p^3}$$
$$\binom{pk}{pa}-\binom{k}{a}=\frac{pk(pk-1)...(pk-pa+1)}{(pa)!}-\frac{k(k-1)...(k-a+1)}{a!}=\binom{k}{a}\Bigg(\frac{\prod_{i=k-a+1}^{k}\binom{pi-1}{p-1}}{\prod_{i=1}^{a}\binom{pi-1}{p-1}}-1\Bigg)$$
Let $$c=\prod_{i=1}^{a}\binom{pi-1}{p-1}$$
Then we have
$$\binom{pk}{pa}-\binom{k}{a}=\frac{\binom{k}{a}}{c}\Bigg(\prod_{i=k-a+1}^{k}\binom{pi-1}{p-1}-\prod_{i=1}^{a}\binom{pi-1}{p-1}\Bigg)$$
Using our lemma, the proof is complete, because
$$\prod_{i=k-a+1}^{k}\binom{pi-1}{p-1}-\prod_{i=1}^{a}\binom{pi-1}{p-1}\equiv1-1\equiv0\pmod{p^3}$$
and
$$c=\prod_{i=1}^{a}\binom{pi-1}{p-1}\equiv 1\pmod{p^3}$$ so $c$ is not divisible by $p$.
Thus, $$\binom{pk}{pa}\equiv\binom{k}{a}\pmod{p^3}$$
For more information, see Ljunggren type congruences.