Rewrite the coverage probability in terms of the MLE: \begin{align*} P_{\theta_0}(\theta_0 \in C_n) = P_{\theta_0}\!\left(|\hat{\theta}_n - \theta_0| \leq \frac{R_n}{\sqrt{n}}\right). \end{align*} Since $R_n \xrightarrow{a.s.} \Phi_0^{-1}(1-\alpha) > 0$, the ratio $\Phi_0^{-1}(1-\alpha)/R_n \xrightarrow{a.s.} 1$. Multiply both sides of the inequality inside the probability by $\Phi_0^{-1}(1-\alpha)/R_n$: \begin{align*} P_{\theta_0}(\theta_0 \in C_n) = P_{\theta_0}\!\left(\frac{\Phi_0^{-1}(1-\alpha)}{R_n} \cdot \sqrt{n}|\hat{\theta}_n - \theta_0| \leq \Phi_0^{-1}(1-\alpha)\right). \end{align*} By the asymptotic normality of the MLE, $\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} N(0, I(\theta_0)^{-1})$. Since $\Phi_0^{-1}(1-\alpha)/R_n \xrightarrow{a.s.} 1$, Slutsky's lemma gives \begin{align*} \frac{\Phi_0^{-1}(1-\alpha)}{R_n} \cdot \sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} N(0, I(\theta_0)^{-1}). \end{align*} Therefore the coverage probability converges to \begin{align*} P\!\left(|Z_0| \leq \Phi_0^{-1}(1-\alpha)\right) = \Phi_0\!\left(\Phi_0^{-1}(1-\alpha)\right) = 1 - \alpha. \qquad \square \end{align*}