Extend the line $QP$ so that it intersects the circle $\Gamma (O,R)$ again at a point which we call $X$. By the sine rule, we obtain \begin{align*} \frac{PM}{PA} &= \sin \angle PAM \\ &= \sin \angle ARB \\ &= \frac{AB}{2R}. \end{align*} Analogously we obtain $\frac{PN}{PB} = \frac{AB}{2R}$. Putting the two together, we get \begin{align*} \frac{1}{PM} + \frac{1}{PN} &= \frac{2R}{AB} \left( \frac{1}{PA} + \frac{1}{PB} \right) \\ &= \frac{2R}{AB} \frac{PA + PB}{PA \cdot PB} \\ &= \frac{2R}{PA \cdot PB}. \end{align*} But the product $PA \cdot PB$ is equal to the power of the point $P$ with respect to $\Gamma$ which is fixed because $P$ is fixed. Hence, the value of $\frac{1}{PM} + \frac{1}{PN}$ does not depend on the choice of chord $AB$. $\square$