Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$
Now if we put $u:=t^{p-1}$ the equation takes the form $$u+\Big( \frac2p -1\Big)\,u^q =\frac {2^{1-\frac p 2 }} p$$ with $q=\frac p{p-1} >1$, that can be solved by series (see e.g. here) (this way one covers an interval $1.57<p\le2$ if I'm not wrong. To cover the other values of $p$, close to $1$, one needs to put the equation in other forms).