Given PQS = 45^ and QPS = 45^.

Given $\angle PQS = 45^\circ$ and $\angle QPS = 45^\circ$.
The sum of angles in a triangle is $180^\circ$.

$$\angle PSQ = 180^\circ - \angle PQS - \angle QPS$$ $$\angle PSQ = 180^\circ - 45^\circ - 45^\circ$$ $$\angle PSQ = 90^\circ$$

This confirms that $PS$ is perpendicular to $QR$.

Since $PS \perp QR$, we have $\angle PSR = 90^\circ$.
We are given $\angle PRT = 30^\circ$. Note that $\angle PRT$ is the same as $\angle PRS$.
So, in right-angled $\triangle PRS$:

$$\angle RPS = 180^\circ - \angle PSR - \angle PRS$$ $$\angle RPS = 180^\circ - 90^\circ - 30^\circ$$ $$\angle RPS = 60^\circ$$

From the diagram, point $T$ lies on the line segment $SR$. Therefore, $\angle RPS$ is the sum of $\angle RPT$ and $\angle SPT$.

$$\angle RPS = \angle RPT + \angle SPT$$

We know $\angle RPS = 60^\circ$ (from Step 2) and we are given $\angle RPT = 30^\circ$.

$$60^\circ = 30^\circ + \angle SPT$$ $$\angle SPT = 60^\circ - 30^\circ$$ $$\angle SPT = 30^\circ$$

The value $QS = 4.5$ cm is not required to solve for $\angle SPT$.

The final answer is $\text{30}^\circ$.