= Solution
Let $A=\operatorname{Leb}(\mathcal C)>0$ and let $\mu$ be the <uniform distribution> on the body. The <hit-and-run kernel density> lower bound implies
$$
P(x,B)\geq\frac{\operatorname{Leb}(B\cap\mathcal C)}{\pi D^2}\geq\varepsilon\mu(B),\qquad
\varepsilon=\frac{A}{2\pi D^2}\in(0,1).
$$
For the last range, $\mathcal C$ lies in a radius-$D$ disc about any of its points, so $A\leq\pi D^2$ and $\varepsilon\leq1/2$. This global <Doeblin condition> proves irreducibility with respect to $\mu$ and aperiodicity. The entire state space is a <small set>; $V=1$, $\lambda=1/2$, and $b=1/2$ satisfy the <geometric drift condition>.
In fact the conclusion is stronger. Write $P=\varepsilon\Pi+(1-\varepsilon)R$, where $\Pi(x,\cdot)=\mu$ and $R$ is the residual <Markov kernel>. Since $\mu P=\mu$, also $\mu R=\mu$. Each transition has a probability $\varepsilon$ of resetting to stationarity, and after such a reset every subsequent distribution remains stationary. Equivalently,
$$
\boxed{\sup_{x\in\mathcal C}\|P^n(x,\cdot)-\mu\|_{\mathrm{TV}}\leq(1-\varepsilon)^n.}
$$
Thus the algorithm has <uniform geometric ergodicity>, using the full-dimensional and boundary conventions stated in part (a).
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