Each interior angle of a regular pentagon is $108^\circ$, so $\angle SRQ=108^\circ$. As $SR=QR$, the triangle is isosceles with $\angle RQS=\angle RSQ = 36^\circ$. Similarly, $\angle SRT= \angle STR = 36^\circ$. So $\angle SUR=(180-2\times36)^\circ=108^\circ$. From the symmetry of the figure, $\angle PUR=\angle PUS= (360^\circ - 108^\circ)/2 = 126^\circ$.

*This problem is taken from the UKMT Mathematical Challenges.*