I'm not a social scientist so anyone with more experience in this field please correct me, but my understanding of the study in question is that the null hypothesis in this case was that socially youth and nontransgender youth have similar levels of anxiety. In this case a higher p value means that the two groups did not show significant differences in their depression which is the conclusion being made.
A quote from the study that this one was following up on:
"In terms of depression, transgender children’s symptoms (M = 50.1) did not differ from the population average, P = .883. In contrast, transgender children had elevated rates of anxiety compared with the population average (M = 54.2), t(72) = 4.05, P < .001"
So the children did not have differing levels of depression, but had statistically significant differences in their levels of anxiety. Basically, this study having high p-values is the reason they conclude that socially accepted transgender youth do not have differing levels of depression from cisgender youth. To throw out the study because of high p-values is to misunderstand the study itself.
The P-value means the margin for error in the conclusions drawn. So a .05 means it's acceptable as scientific evidence. Anything above that makes it unacceptable. You can still draw conclusions from it, but it will not be taken as scientific evidence unless the p-value is lower than .05.
The study has such a bad p-value because the sample size is so small. Only like 50 patients. Even a sample of 1000 patients would be iffy by these standards and may not pass the p-value test. So yea, the study doesn't mean anything.
In fact, the conclusions with the best p-value are the conclusions that favor the conservative point of view. "Trans reported marginally higher anxiety compared to the control group p=0.076" and "trans reported marginally higher anxiety than the national average =0.096" and finally, the only scientifically relevant fact present "parents reported their children had more anxiety than children in the control groups =0.002"