Conside the line from E to RC that is parallel to BC. Let the point of intersection be X

EX=24

XR=24-6=18

find ER

\(ER=\sqrt{24^2+18^2}\\ ER=\sqrt{900}\\ ER=30 \)

Now PR is just the diagonal of one side of the square

\(PR=\sqrt{24^2+24^2}\\ PR=\sqrt{1152}\\ \)

Now I have all 3 sides of the triangle so I can use cosine rule to find the angle between them

the sides are \(30, \quad30, \quad \sqrt{1152}\)

\(1152=900+900-2*900*Cos(PER)\\ 1152=1800-1800*Cos(PER)\\ -648=-1800*Cos(PER)\\ 0.36=Cos(PER)\\ \angle{PER}=cos^{-1}0.36=68.8998^\circ=68^\circ54' \)

That is assuming I've not made any stupid mistakes.