直线$x-ky-1=0$过定点$\left(1,0\right)$，可得以$OP$为直径的圆的方程为：${(x-\frac{1}{2})}^{2}+{y}^{2}=\frac{1}{4}$，设$x_{0}+y_{0}=b$，即$x_{0}+y_{0}-b=0$，又$M$在$OP$为直径的圆：${(x-\frac{1}{2})}^{2}+{y}^{2}=\frac{1}{4}$上，即$(x_{0}$，$y_{0})$是$x+y-b=0$与圆${(x-\frac{1}{2})}^{2}+{y}^{2}=\frac{1}{4}$的交点，$\therefore \frac{|\frac{1}{2}-b|}{\sqrt{2}}≤\frac{1}{2}$，$b≤\frac{\sqrt{2}+1}{2}$.故选：$C$.