With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as $$G(r,y):=2^r y^{2 r-1} (r+y)-1=0. \tag{2}$$ For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.

Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)>0$, for each $r$. So, for each $r$, equation (2) has a unique root $Y(r)\in(0,1/2)$. In fact, for each $r$ we have $G(r,4/10)>0$. So, for each $r$ $$Y:=Y(r)\in(0,4/10). \tag{3}$$ Moreover, $$G'_y(r,y)=2^r r y^{2 r-2} (2 r+2 y-1)>0.$$ So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover, $$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y}\overset{\text{sign}}=-(r+Y) \ln \left(2 Y^2\right)-1>-(1/2+Y) \ln \left(2 Y^2\right)-1>0,$$ the latter inequality holding in view of (3); of course, $\overset{\text{sign}}=$ means the equality in sign: $a\overset{\text{sign}}=b$ means $\text{sign}\, a=\text{sign}\,b$.

Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.

With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as $$G(r,y):=2^r y^{2 r-1} (r+y-r y)-1=0. \tag{2}$$ For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.

Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)=2^{-r} (1+r - 2^r)>0$, for each $r$; here one may use the concavity of $1+r - 2^r$ in $r$. So, for each $r$, equation (2) has a unique root $$Y:=Y(r)\in(0,1/2). \tag{3}$$ Moreover, $$G'_y(r,y)=2^{r+1} r y^{2 r-2}(r-1/2 + (1-r)y)>0.$$ So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover, $$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y} \overset{\text{sign}}=H(r,Y)>H(1/2,Y)\overset{\text{sign}}=h(Y), $$ where $a\overset{\text{sign}}=b$ means $\text{sign}\, a=\text{sign}\,b$, $H(r,y):=-1 + y - (y + r (1 - y)) \ln(2 y^2)$, and $$h(y):= -\frac{1 - y}{1 + y} - \frac12\,\ln(2 y^2).$$ Note that $h(1/2)>0$ and $h'(y)=-\frac{1+y^2}{y (1+y)^2}<0$, whence $h>0$ and hence $Y'>0$.

Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.

With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as $$G(r,y):=2^r y^{2 r-1} (r+y)-1=0. \tag{2}$$ For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.

Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)>0$, for each $r$. So, for each $r$, equation (2) has a unique root $Y(r)\in(0,1/2)$. In fact, for each $r$ we have $G(r,4/10)>0$. So, for each $r$ $$Y:=Y(r)\in(0,4/10). \tag{3}$$ Moreover, $$G'_y(r,y)=2^r r y^{2 r-2} (2 r+2 y-1)>0.$$ So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover, $$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y}\overset{\text{sign}}=-(r+Y) \ln \left(2 Y^2\right)-1>-(1/2+Y) \ln \left(2 Y^2\right)-1>0,$$ the latter inequality holding in view of (3); of course, $\overset{\text{sign}}=$ means the equality in sign.

Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.

With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as $$G(r,y):=2^r y^{2 r-1} (r+y)-1=0. \tag{2}$$ For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.

Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)>0$, for each $r$. So, for each $r$, equation (2) has a unique root $Y(r)\in(0,1/2)$. In fact, for each $r$ we have $G(r,4/10)>0$. So, for each $r$ $$Y:=Y(r)\in(0,4/10). \tag{3}$$ Moreover, $$G'_y(r,y)=2^r r y^{2 r-2} (2 r+2 y-1)>0.$$ So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover, $$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y}\overset{\text{sign}}=-(r+Y) \ln \left(2 Y^2\right)-1>-(1/2+Y) \ln \left(2 Y^2\right)-1>0,$$ the latter inequality holding in view of (3); of course, $\overset{\text{sign}}=$ means the equality in sign: $a\overset{\text{sign}}=b$ means $\text{sign}\, a=\text{sign}\,b$.

Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.