Let $F$ be the non-archimedean local field $\mathbb{Q}_p$ for some prime $p$ and $D$ be a quaternion division algebra over $F$. Let $\mathcal{O}_D$ and $\mathcal{P}_D$ denote the ring of integers of $D$ and its unique maximal ideal (respectively). Then, what is the finite group $$ \frac{D^*}{F^*(1+ \mathcal{P}_D)} = {?}$$ where $D^*=D-\{0\}$ and $F^*=F-\{0\}$ are multiplicative groups.

Consider the reduced norm map $N_\text{rd}:D \rightarrow F$, then $N_\text{rd}(D^*)=F^*$ and if $D^1$ denotes the reduced norm one elements of $D$, then we have an exact sequence $$1 \rightarrow D^1 \rightarrow D^* \rightarrow F^* \rightarrow 1$$ but we have $D^1 \cap F^*=\{\pm 1\}$. We know from Carl Riehm's article that $$ \frac{D^1}{(1+ \mathcal{P}_D)} \cong {_N}(\mathbb{F}_{p^2})= \text{Finite cyclic group of order } (p+1).$$ Here ${_N}(\mathbb{F}_{p^2})$ is the subgroup of $\mathbb{F}_{p^2}$ consisting of norm 1 elements.

Question: Similarly, can we write $ \frac{D^*}{F^*(1+ \mathcal{P}_D)}$ in terms of finite fields?

Any comments or suggestions will be extremely helpful.