Lower bound on the worst-case unbiased coin flips to sample a distribution?
Fix such a distribution, and an element $k$ showing its non-triviality. In particular, $\Pr[X=k]= (2m+1)/2^p$ for some integer $m$.

Suppose by contradiction you have an algorithm A to sample from the distribution given $p-1$ i.i.d. unbiased bits. Such an algorithm is, given the realization of the coins, deterministic: i.e., it is equivalent to a partition of the leaves of the complete binary tree of height $p-1$, each output $x$ corresponding to a different set $S_x$ in the partition.

As such, since all leaves have probability $1/2^{p-1}$, any output $x$ has probability $|S_x|/2^{p-1}$, i.e., of the form $2\ell/2^p$ for some integer $\ell$. That cannot be equal to $(2m+1)/2^p$.