Solution:

a) We shall prove that the position of the point $S = O_{1}O_{2} \cap AB$ does not depend on $k$. Let $O_{3}$ be the center of $k$. It follows by the Menelaus theorem for $\triangle O_{1}O_{2}O_{3}$ and the line $AB$ that
$$\frac{O_{3}B}{BO_{2}} \cdot \frac{O_{2}S}{SO_{1}} \cdot \frac{O_{1}A}{AO_{3}} = 1$$
Since $O_{3}B = AO_{3}$, we get $\frac{O_{2}S}{SO_{1}} = \frac{BO_{2}}{O_{1}A} = \frac{R_{2}}{R_{1}} = \frac{16}{4} = 4$.
Hence $S$ is a fixed point and the equalities $O_{1}O_{2} = 25 = O_{2}S + O_{1}S$ imply that $O_{2}S = 20$ and $O_{1}S = 5$.

b) Setting $\angle O_{1}O_{3}O_{2} = x$ and $\angle O_{1}O_{2}O_{3} = y$, then $\angle AO_{1}S = x + y$. Since $\triangle O_{1}AP$ is isosceles, we have $\angle APS = \frac{x + y}{2}$. On the other hand, the triangles $AO_{3}B$ and $BO_{2}Q$ are also isosceles; hence $\angle SBO_{3} = 90 - \frac{x}{2}$ and $\angle QBO_{2} = 90 - \frac{y}{2}$, which implies that $\angle SBQ = \frac{x + y}{2}$. Therefore $\angle APS = \angle SBQ$, i.e. $PBQA$ is a cyclic quadrilateral.

c) Note that $SP \cdot SQ = SA \cdot SB$ and $SP \cdot SQ = (SO_{1} + R_{1})(SO_{2} - R_{2}) = 9 \cdot 4 = 36$. The inequality
$$AB = SA + SB \geq 2 \sqrt{SA \cdot SB} = 2 \sqrt{SP \cdot SQ} = 12$$
implies that the minimum of $AB$ equals $12$ and it is attained if $SA = SB$.
It remains to show that there is a circle $k$ with $SA = SB$. Take a point $A \in k_{1}$ such that $SA = 6$. Since the power of $S$ with respect to $k_{1}$ equals $SO_{1}^{2} - R_{1}^{2} = 5^{2} - 4^{2} = 9$ and $SA^{2} = 36 > 9$, it is easy to see that there is a circle $k$ passing through $A$ and satisfying the conditions of the problem. Then $SB = \sqrt{SP \cdot SQ} = 6$, i.e., $SA = SB$.