Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with projections $p,q\in\mathcal{A}$ and a state $\varphi:\mathcal{A}\to \mathbb{C}$. Consider $p\wedge q$ in the bidual $\mathcal{A}^{\ast\ast}$. Where $\omega_\varphi$ is the extension of $\varphi$ to a normal state on $\mathcal{A}^{\ast\ast}$, I believe the following to be (almost obviously(!)) true:
Proposition: $\omega_\varphi(p\wedge q)=1$ if and only if $\varphi(p)=\varphi(q)=1$.
The issues I potentially have with the proof are around some technical issues. I will try and flag these with (!!) and hopefully there are no others. I would also be happy to see other proofs.
Proof: Assume that $\omega_\varphi(p\wedge q)=1$. Let $\widetilde{\pi_\varphi}$ be the GNS representation of $\omega_\varphi:\mathcal{A}^{\ast\ast}\to\mathbb{C}$ with cyclic vector $\widetilde{x}\in\widetilde{\mathsf{H}_\varphi}$. As $\widetilde{\pi_\varphi}$ is strongly continuous (!!1), and $\operatorname{sot-lim}_{n\to \infty}((pq)^n)=p\wedge q$: $$\widetilde{\pi_{\varphi}}(p\wedge q)=\widetilde{\pi_{\varphi}}(p)\wedge\widetilde{\pi_{\varphi}}(q)\qquad(\text{!!2}).$$ Now, $$\omega_{\varphi}(p\wedge q)=1 \implies\langle \widetilde{x},\widetilde{\pi_{\varphi}}(p\wedge q)\widetilde{x}\rangle=1\implies \widetilde{x}\in \operatorname{ran}(\widetilde{\pi_\varphi}(p\wedge q))$$ and because of (!!2), this implies $$\widetilde{x}\in \operatorname{ran}(\widetilde{\pi_\varphi}(p))\cap \operatorname{ran}(\widetilde{\pi_\varphi}(p)).$$ This implies that, where $p,q\in\mathcal{A}^{\ast\ast}$ via the embedding $\mathcal{A}\subseteq \mathcal{A}^{\ast\ast}$, $\omega_\varphi(p)=\omega_\varphi(q)=1$ which implies that $\varphi(p)=\varphi(q)=1$.
Now suppose $\varphi(p)=\varphi(q)=1$ and let $\pi_\varphi$ be the GNS representation of $\varphi$ with cyclic vector $x\in\mathsf{H}_{\varphi}$ so that $$\varphi(f)=\langle x,\pi_\varphi(f)x\rangle.$$ That $\varphi(p)=\varphi(q)=1$ implies that $x\in \operatorname{ran} \pi_{\varphi}(p)\cap \operatorname{ran}\pi_{\varphi}(q)$. Now: $$\omega_\varphi(p\wedge q)\underset{\text{!!3}}{=}\lim_{n\to \infty}\varphi((pq)^n)=\lim_{n\to \infty}\langle x,\pi_\varphi((pq)^n)x\rangle=\lim_{n\to \infty}\langle x,(\pi_\varphi(p)\pi_\varphi(q))^nx\rangle=1,$$ as required.
All your issues are solved from the fact that $\omega_\varphi$ is normal state on a von Neumann algebra, and hence its GNS representation is normal, which is to say that $\tilde\pi_\varphi$ is strongly continuous on bounded sets; which is exactly what you need.
That said, this is a really weird way of doing the proof. You have that $p\wedge q\leq p$ and you are dealing with states, so $$ 1=\omega_\varphi(p\wedge q)\leq \omega_\varphi(p)=\varphi(p)\leq 1, $$ giving you $\varphi(p)=1$.