Let $0<p<1$, $\lambda>0$, $x_1 \neq 0$, $x_2\neq 0$, $x_1 - c_1 \neq 0$, $x_2 - c_2 \neq 0$, $\xi>0$. Also, either $c_1\neq 0$ or $c_2 \neq 0$ but both cannot be zero simultaneously. Can we show the following system of nonlinear equations has finitely many solutions?: $$ \begin{aligned} p\,\text{sign}(x_1)|x_1|^{p-1}+\lambda(x_1-c_1)&=0\\ p\,\text{sign}(x_2)|x_2|^{p-1}+\lambda(x_2-c_2)&=0\\ (x_1-c_1)^2+(x_2-c_2)^2 &= \xi^2 \end{aligned} $$
My try: From the first and second equations we get the following : $$ \frac{p^2(|x_1|^{2(p-1)}+|x_2|^{2(p-1)})}{\lambda^2}=\xi^2 $$ where squaring the terms eliminates $\text{sign}(x_1)$ and $\text{sign}(x_2)$. Since $x_1$ and $x_2$ are on a circle, they are bounded. Also, according to our assumptions, they cannot be zero so the numerator of the left hand side is bounded which implies $\lambda$ should be bounded.
Note:
For rational number specially $p=1/2$, it can be shown that it has finitely many solutions. I am interested in the non restricted case where $p$ can take irrational numbers as well.
I guess that you assume that $c_1$, $c_2$ (and not $x_1-c_1$, $x_2-c_2$) are non null, and that the unknowns are $x_1$ and $x_2$?
For all real number $x$, set $f(x) = p~\mathrm{sign}(x)~|x|^{p-1} + \lambda x$. Then $f(x)$ has the same sign as $x$. When $x \ne 0$, $f'(x) = p(p-1)|x|^{p-2} + \lambda$ has the same sign as $|x|^{p-2} - \lambda/[p(1-p)]$, namely the same sign as $x_0-|x|$, where $$x_0 := \Big(\frac{\lambda}{p(1-p)}\Big)^{1/(p-2)}.$$ Therefore $f$ is increasing on $(-\infty,-x_0]$ and on $[x_0,+\infty)$ and decreasing on $[-x_0,0)$ and on $(0,x_0]$.
As a result, the first equation has at most two solutions $x_1$, the second equation has at most two solutions $x_2$, so the system has at most four solutions $(x_1,x_2)$.