I think that in this case there will always be a complete bunch $B$.
- For every maximal chain $m$ with $m \cap t = \varnothing$, take the minimal element of $m$, and add it as tods to $B$.
- If $m \cap t \ne \varnothing$, and $t \not\subset m$, then take the minimal element $x$ of $m \setminus t$, and add $\{x\} \cup (m \cap t)$ to $B$.
- Finally, add $t$ to $B$, and now $B$ is complete.
$B$ is also a bunch of finite tods, because by construction all tods in $B$ are finite and pairwise incompatible.