If $T$ is sufficient for $\theta$, let $\lambda$ be an LRT statistic for $\mathbf{X}$ and $\lambda^*$ and LRT statistic for $T$. Is it true that $\lambda=\lambda^*$?
According to theorem 8.2.4 of the book "Statistical Inference" by Casella and Berger, seems to be the case, but the proof relies on an artifact of the factorization theorem that I think is only true with discrete distributions. The proof is as follows \begin{equation} \begin{aligned} \lambda(\mathbf{x})&:=\frac{\sup_{\Theta_0}L_{\mathbf{x}}(\theta)}{\sup_{\Theta}L_{\mathbf{x}}(\theta)}=\frac{\sup_{\Theta_0}f_{\theta}(\mathbf{x})}{\sup_{\Theta}f_{\theta}(\mathbf{x})}\\&=\frac{\sup_{\Theta_0}g_{\theta}(T(\mathbf{x}))h(\mathbf{x})}{\sup_{\Theta}g_{\theta}(T(\mathbf{x}))h(\mathbf{x})}=\frac{\sup_{\Theta_0}g_{\theta}(T(\mathbf{x}))}{\sup_{\Theta}g_{\theta}(T(\mathbf{x}))}=\lambda^*(T(\mathbf{x}))\text{, } \end{aligned} \end{equation} according to the book, the last equality is true as long as $g_{\theta}$ is the density function of $T$, but I think this only holds if the population is discrete (watch this).
The test you showed is the Generalized one...but just to understand the problem take as an example the more suitable Neyman Pearson Lemma for simple hypothesis.
You have that the test is
$$\frac{L(\theta_0|\mathbf{x})}{L(\theta_1|\mathbf{x})}\leq k$$
now observe that "IF" a sufficient estimator $t(\mathbf{x})$ esists NP lemma becomes
$$\frac{h(\mathbf{x})\cdot g(t(\mathbf{x}),\theta_0)}{h(\mathbf{x})\cdot g(t(\mathbf{x}),\theta_1)}\leq k$$
$$\frac{ g(t(\mathbf{x}),\theta_0)}{ g(t(\mathbf{x}),\theta_1)}\leq k$$
thus the rejection region depends by the data ONLY through a Statistic, that is $t(\mathbf{x})$, sufficient for the model.
Example 1:
$X\sim Exp(\theta)$
You perform the test using $T=\Sigma_iX_i$, sufficient statistic (in this case, Minimal, Sufficient and Complete) and you can use chi-squared tables,
Example 2
$X\sim N(\mu;1)$
you perform your test again using $T=\Sigma_iX_i$ or, what is the same, the sample mean...again $T$ is CSS (complete and sufficient)
Example 3
$X\sim U(0;\theta)$
and you perform your test basing it on $T=X_{(n)}$, the maximum of your observations, Complete and sufficient statistics
Other UMP test generalizing NP Lemma or Generalized LR Test (which is not anymore UMP) work in the same way: if a Sufficient estimator exists, the test is based on this important estimator... and it is very intuitive because the Sufficient estimator is so named as it include all the information you need to estimate the parameter...