Draw the triangle PQR. Then draw a segment (PD) from P to QR such that RS is congruent to PR. From the base angles theorem you know that m<RDP = m<RPD. Because m<P = m<RPD + m<DPQ, you know that m<P = m<RDP + m<DPQ. Therefore m<P > m<RDP.

From the base angles theroem, you know that m<RDP = m<DPC + m<Q and therfore that m<RDP > m<Q (this idea is the exterior angle inequality theorem). Finally, because m<P > m<RDP and m<RDP > m<Q, you can conclude m<P > m<Q.

So there is the paragraph proof. I'll leave it for you to put the statements and reasons into a two-column format.