The inequality $\widetilde{P}_{\theta}(u)\leq P_{\theta}(u)$ is trivial. To prove that equality holds on $X^{\mathrm{qm}}$, pick $\varepsilon>0$ and $x\in\emb_{\mathcal{X}}(\Delta_{\mathcal{X}})$ for some SNC model $\mathcal{X}$ on which $\theta$ is determined. By construction, there exists $\psi\in\PSH(X,\theta)$ such that $\psi\leq 0$ and $\psi(x)\geq P_{\theta}(u)(x)-\varepsilon$. By the definition of $\PSH(X,\theta)$, there then exists a $\theta$-psh model function $\varphi$ such that $|\varphi-(\psi-\varepsilon)|\leq\varepsilon$ on $\emb_{\mathcal{X}}(\Delta_{\mathcal{X}})$. Thus $\varphi\leq\varphi\circ p_{\mathcal{X}}\leq\psi\circ p_{\mathcal{X}}\leq 0$ on $X$ and $\varphi(x)\geq\psi(x)-2\varepsilon\geq P_{\theta}(u)(x)-3\varepsilon$. We conclude that $\widetilde{P}_{\theta}(u)=P_{\theta}(u)$ on $X^{\mathrm{qm}}$. ∎