Solution:

Let $(c_1)$ and $(c_2)$ be circles through $K$, internally tangent to $(c)$ at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $(c_1)$ and $(c)$ meet the common tangent to $(c_2)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $(c_1)$, $(c_2)$ and $(c)$, and the line $KL$ passes through $Q$.

We have $\text{m( QLB ) = m( QBK ) = m( QBA ) = 1 2 m( BA ) = m( QOB )}$. So, the quadrilateral $OBQL$ is cyclic. We conclude that $m(\widehat{QLO}) = 90^{\circ}$ and the points $O, B, Q, A$ and $L$ are cocyclic on a circle $(k)$.

From $MO^2 - OP^2 = \mathcal{P}_c(M) = MA \cdot MB = \mathcal{P}_k(M) = ML \cdot MO = (MO - OL) \cdot MO = MO^2 - OL \cdot MO$ follows that $OP^2 = OL \cdot OM$. Since $PL \perp OM$, this shows that the triangle $MPO$ is right at point $P$. Thus, the line $MP$ is tangent to the circle $(c)$.

Let $R \in (c)$ be so that $BR \perp MO$. The triangle $LBR$ is isosceles with $LB = LR$, so $\widehat{OLR} \equiv \widehat{OLB} \equiv \widehat{OQB} \equiv \widehat{OQA} \equiv \widehat{MLA}$. We conclude that the points $A, L$ and $R$ are collinear.

Now $m(\widehat{AMR}) + m(\widehat{AOR}) = m(\widehat{AMR}) + 2 m(\widehat{ABR}) = m(\widehat{AMR}) + m(\widehat{ABR}) + m(\widehat{MRB}) = 180^{\circ}$, since the triangle $MBR$ is isosceles. So, the quadrilateral $MAOR$ is cyclic.

This yields $LM \cdot LO = -\mathcal{P}_{(MAOR)}(L) = LA \cdot LR = -\mathcal{P}_c(L) = LP^2$, which as above, shows that $OP \perp PM$.

$\widehat{KLA} \equiv \widehat{KAQ} \equiv \widehat{KLB}$ and $m(\widehat{MLK}) = 90^{\circ}$ show that $[LK$ and $[LM$ are the internal and external bisectors of the angle $\widehat{ALB}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So, $LK$ is the polar line of $M$ in the circle $(c)$.