= Solution
Put $r(\theta)=p(\mathcal D\mid\theta)p(\theta)$ and $Z=\int r(\theta)\,d\theta$. The <support condition for importance sampling> is essential: $q(\theta)>0$ almost everywhere that $r(\theta)>0$. For one draw $\theta\sim q$, the <importance weight> $W=r(\theta)/q(\theta)$ satisfies
$$
\mathbb E_q W=\int_{\{q>0\}}\frac{r(\theta)}{q(\theta)}q(\theta)\,d\theta=Z.
$$
By <linearity of expectation>,
$$
\boxed{\mathbb E\widehat p_1(\mathcal D)=\frac1T\sum_{t=1}^T\mathbb E_qW_t=Z=p(\mathcal D)}.
$$
This proves that it is an <unbiased estimator>. If the proposal omits a region of positive target mass, the <expected value> instead misses that region's contribution; the unconditional unbiasedness claim needs the stated <support condition for importance sampling>.
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