I'm going to call manifolds with at most $L$ simplices incident to a vertex "of geometry bounded by $L$" or just "of bounded geometry". The operator $*$ denotes the join.
First, using the Delaunay triangulation argument in Ian Agol's answer, every PL sphere of geometry bounded by $L$ can be filled with a PL ball of geometry bounded by $C(L)$. The procedure is: smooth it out, take a Riemannian filling, take a Delaunay triangulation of the filling, and then use some finite procedure to interpolate between the original boundary and the new boundary which may have gotten subdivided. (This is the shakiest part of the argument.)
Now take a PL manifold $M^d$ equipped with an arbitrary PL triangulation.
- Barycentrically subdivide once. The resulting complex has a vertex for each face of $M$. The link of the vertex corresponding to a $k$-face $\sigma$ looks like $\partial\Delta^{d-k} * \operatorname{BarySub}(\operatorname{link}(\sigma))$. The star of the vertex corresponding to every $d$-face and $(d-1)$-face already has geometry bounded by $C(d)$.
- Now look at the star of each vertex corresponding to a $(d-2)$-face. This looks like $\partial\Delta^{d-2} * CS$ where $S$ is a circle with many edges. We can replace the interior of this subcomplex by $\partial\Delta^{d-2} * D(S)$, where $D(S)$ is a bounded geometry filling of $S$. Now this subcomplex has geometry bounded by $C(d,2)$.
- If we now look at the star of each vertex corresponding to a $(d-3)$-face, it looks like $\partial\Delta^{d-3} * C\Sigma$ where $\Sigma$ is a surface. This surface is a union of patches which are the $D(S)$'s from the previous step, therefore it has bounded geometry. We can replace $C\Sigma$ with a 3-ball of geometry bounded by $C(d,3)$.
- Keep going like this. At the end of the procedure, the resulting manifold has bounded geometry.