This is my effort in solving the problem.
Here is my work:
[math]S\widehat{Q}R=44\degree\ ~(\text{>s in alternate segment})[/math]
[math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]S\widehat{Q}R= Q\widehat{S}R=44\degree ~(\text{base >s of an isosceles triangle})[/math]
[math]Q\widehat{R}S=180-(2\times 44) = 92\degree~ (\text{sum of >s in a triangle })[/math]
[math]Q\widehat{R}S= Q\widehat {P}S=92\degree ~(\text{opposite >s of a cyclic quadrilateral})[/math]
[math]P\widehat{S}Q=180-(92+31) = 57\degree[/math]
Since [math]P\widehat{S}R=P\widehat{S}Q + Q\widehat{S}R[/math]
[math]P\widehat{S}R =57\degree +44\degree =101\degree[/math]
Is my solution correct?