Пусть АК и BР биссектрисы, которые пересекаются в точке О.
\(\angle \,PAO = \angle \,BAO = \alpha ;\,\,\,\,\,\,\,\,\,\angle \,ABO = \angle \,KBO = \beta \,\,\,\,\)
Из треугольника ABС:
\(2\alpha + 2\beta + {90^ \circ } = {180^ \circ }\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,2\alpha + 2\beta = {90^ \circ }\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\alpha + \beta = {45^ \circ }\).
Из треугольника AOB:
\(\angle \,AOB + \alpha + \beta = {180^ \circ }\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\angle \,AOB + {45^ \circ } = {180^ \circ }\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\angle \,AOB = {135^ \circ }\).
Тогда: \(\angle \,AOP = {180^ \circ } — {135^ \circ } = {45^ \circ }\)
Ответ: 45.