Finding root of symbolic function

The t/s is somehow a nasty in this question. The inequality chain make t/s limited by 0 and 1. 1 means t==s. At this point the equation reduces to Sqrt[1-p] and is satisfied only for p==1.

Following this path replace u=t/s and solve the equation.

p[theta_, u_] := 1 - (theta^u - theta)^2

Plot3D[p[theta, u], {theta, 0, 1}, {u, 0, 1}, AxesLabel -> Automatic]

The quotient t/s makes the equation unnecessary complicated. u makes it trivial.

It is too meaningful to replace the Sqrt[1-p] by q for example and

Plot3D[theta^u - theta, {u, 0, 1}, {theta, 0, 1}]

For very large s the quotient t/s->0 and the equation gets

1-theta-Sqrt[1-p]

Reduce[1 - theta - Sqrt[1 - p] == 0, {p, theta}]

(* theta == 1 - Sqrt[1 - p] *)

Plot3D[theta == 1 - Sqrt[1 - p], {p, 0, 1}, {theta, 0, 1}, 
 AxesLabel -> Automatic, PlotRange -> All]

Reduce[theta^u - theta - Sqrt[1 - p] == 0 && theta > 0 && theta < 1 &&
   0 < p < 1, u]
(* Element[C[1], Integers] && 0 < p < 1 && 0 < theta < 1 && 
  u == (2*I*Pi*C[1] + Log[Sqrt[1 - p] + theta])/Log[theta] *)

Plot3D[Log[Sqrt[1 - p] + theta]/Log[theta], {p, 0, .95}, {theta, 0, .9}, 
 AxesLabel -> Automatic, PlotRange -> All]

This is not so informativ and comfortable to look at if p->1 and theta->1.