On the event $\hat{\theta}_n \in \text{int}(\Theta)$ (which occurs with probability tending to 1), expand the log-likelihood around $\hat{\theta}_n$ using a second-order Taylor expansion. Since $\nabla_\theta \ell_n(\hat{\theta}_n) = 0$ at the MLE, the expansion of $\ell_n(\theta_0)$ around $\hat{\theta}_n$ gives \begin{align*} \Lambda_n = 2\ell_n(\hat{\theta}_n) - 2\ell_n(\theta_0) = \sqrt{n}(\theta_0 - \hat{\theta}_n)^\top \bar{B}_n \sqrt{n}(\theta_0 - \hat{\theta}_n), \end{align*} where $\bar{B}_n$ is defined entry-wise by $(\bar{B}_n)_{ij} = -\frac{\partial^2}{\partial\theta_i\partial\theta_j}\bar{\ell}_n(\bar{\theta})$ for some $\bar{\theta}$ on the segment $[\theta_0, \hat{\theta}_n]$. By uniform convergence of the Hessian, $\bar{B}_n \xrightarrow{\mathbb{P}_{\theta_0}} I(\theta_0)$. Since $\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} N(0, I(\theta_0)^{-1})$, the quantity $I(\theta_0)^{1/2}\sqrt{n}(\hat{\theta}_n - \theta_0) \xrightarrow{d} N(0, I_p)$, and thus $\Lambda_n \xrightarrow{d} \chi^2_p$ by Slutsky's lemma.