In PQS, R is the mid-point of QS.

Since $PR = QR$, $\triangle PQR$ is an isosceles triangle. The angles opposite the equal sides are equal.

$$\angle QPR = \hat{Q} = 56^\circ$$

The sum of angles in a triangle is $180^\circ$. In $\triangle PQR$:

$$\hat{R}_1 + \angle QPR + \hat{Q} = 180^\circ$$ $$\hat{R}_1 + 56^\circ + 56^\circ = 180^\circ$$ $$\hat{R}_1 + 112^\circ = 180^\circ$$ $$\hat{R}_1 = 180^\circ - 112^\circ$$ $$\hat{R}_1 = \boxed{\text{68}^\circ}$$

Angles $\hat{R}_1$ and $\hat{R}_2$ (which is $\angle PRS$) are supplementary angles because they form a straight line at R on QS.

$$\hat{R}_1 + \hat{R}_2 = 180^\circ$$ $$68^\circ + \hat{R}_2 = 180^\circ$$ $$\hat{R}_2 = 180^\circ - 68^\circ$$ $$\hat{R}_2 = 112^\circ$$

Since R is the mid-point of QS, $QR = RS$. We are given $PR = QR$.
Therefore, $PR = RS$. This means $\triangle PRS$ is an isosceles triangle.
The angles opposite the equal sides are equal:

$$\angle RPS = \hat{S}$$

The sum of angles in $\triangle PRS$ is $180^\circ$:

$$\hat{R}_2 + \angle RPS + \hat{S} = 180^\circ$$ $$112^\circ + \hat{S} + \hat{S} = 180^\circ$$ $$112^\circ + 2\hat{S} = 180^\circ$$ $$2\hat{S} = 180^\circ - 112^\circ$$ $$2\hat{S} = 68^\circ$$ $$\hat{S} = \frac{68^\circ}{2}$$ $$\hat{S} = \boxed{\text{34}^\circ}$$