<Tri-2024-2105>,hint

Suppose circumcircle of triangle MCQ  cuts AC and AM at C and T ,respectively

Since <QCC'=<QC'C=90-C. Hence CC'=2 MC sin (90-C)=2R sinA cosC

AC'=AC-CC'=2RsinB -2R sinA cosC=(2R sin(A+C)-2RsinA cosC)=

2R (sinAcosC +cosA sinC-sinA cosC)=2R sinC cosA. 

Hence AC xAC'=2R sinB x 2R sinC cosA=AT x AM, <MTQ=90

Suppose circumcircle of triangle MBP cuts  AB and AM at B' and T'

Similarly AB x AB'=2R  x 2R cosA sinB sinC=AC x AC'

Hence AT=AT'. <MT'P=90. Hence AM and PQ are perpendicular. A,T,Q are collinear.