$X$ is an unbiased estimator of $p$, & is indeed sufficient, but not minimal sufficient: $$\Pr(X=2|X\in\{2, 3\})= \Pr(X=3|X\in\{2, 3\}) = \tfrac{1}{2} $$ regardless of the true value of $p$; & so a minimal sufficient statistic $T$ partitions the sample space into $X=1$ & $X\in \{2,3\}$. Then if you want an unbiased estimator with lower variance than $X$, Rao–Blackwellize: $$\operatorname{E}{X|T}= \begin{cases} 1\text{ when }X=1\\ 2\tfrac{1}{2} \text{ when } X\in\{2,3\}\end{cases}$$ $$\operatorname{E}{X|T}= \begin{cases} 1 & \text{ when }X=1\\ 2\tfrac{1}{2} & \text{ when } X\in\{2,3\}\end{cases}$$ This is about as 'standard' as inference gets. (Though perhaps more typical is the case where an unbiased estimator is not sufficient, & can be improved by Rao–Blackwellizing it—say, with a sample size greater than 1, $\bar X$ as an estimator of $p$.)
Note that $T$ may be coded as '0' & '1', for $X=1$ & $X\in\{2,3\}$ respectively; & then writing $\pi=\Pr(T=1)=\tfrac{2(p-1)}{3}$ makes it apparent that we've been concerned with inference about the probability parameter of a single Bernoulli trial. If anything still seems unintuitive, it's perhaps because you'd typically formulate such a model straight off, without drawing unnecessary distinctions between evidentially equivalent outcomes. (Suppose parts coming off an assembly line are tested, & an unknown proportion rejected; then half of those that aren't rejected are painted red and the other half painted blue.)