Sum involving binomial coefficient satisfies congruence (A contest question)

Here, I think, is an idea in the right direction. Consider the formal power series expansion

$$F(x) := \frac{1}{\sqrt{1-4x}} = \sum_{k=0}^\infty {2k \choose k} x^k.$$

View this identity in $\mathcal O[[x]]$, the ring of formal power series with coefficients in the ring of integers $\mathcal O$ of $\overline{\mathbf Q_p}$. Let $i \in \mathcal O$ denote a square root of $-1$. Multiplying both numerator and denominator of $$\frac{1}{\sqrt{1-4x}}$$ by $i$, we have

$$F(x) = \frac{i}{\sqrt{4x-1}} = \frac{i}{\sqrt{1-4(\frac12-x)}} = iF(\frac12-x)$$

Reducing modulo the maximal ideal of $\mathcal O$, and using the fact that the central binomial coefficients are eventually divisible by $p$, we find that $$-\overline i g(x) =g\left(\frac12 - x\right) \text{ in }\overline{\mathbf F_p}$$

where $$g(x) = \sum_{k=0}^{2p-1}{2k \choose k} x^k.$$

This isn't what the question asks, but it's similar enough in appearance to lead me to believe that this is not a coincidence!