1. Identify Key Points and Angles:
- Let $M$ be the midpoint of $AC$ in the acute-angled triangle $ABC$ with $AB > BC$.
- Let $\Omega$ be the circumcircle of $\triangle ABC$.
- The tangents to $\Omega$ at points $A$ and $C$ meet at $P$.
- $BP$ and $AC$ intersect at $S$.
- Let $AD$ be the altitude of $\triangle ABP$.
- Let $\omega$ be the circumcircle of $\triangle CSD$.
- Suppose $\omega$ and $\Omega$ intersect at $K \neq C$.

2. Establish Concyclic Points:
- Since $K$ lies on $\omega$, we have $\angle PDK = 180^\circ - \angle SDK = \angle KCS$.
- Since $K$ lies on $\Omega$, we have $\angle AKC = 180^\circ - \angle ABC$.
- Since $PA$ is tangent to $\Omega$, we have $\angle PAC = \angle ABC$.

3. **Calculate $\angle KCS$:**
$$\angle KCS = 180^\circ - \angle AKC - \angle KAC$$
$$\angle AKC = 180^\circ - \angle ABC$$
$$\angle PAC = \angle ABC$$
$$\angle KCS = 180^\circ - (180^\circ - \angle ABC) - (\angle PAC - \angle PAK)$$
$$\angle KCS = \angle ABC - \angle PAC + \angle PAK$$
$$\angle KCS = \angle PAK$$

4. Conclude Concyclic Points:
- Thus, $\angle PDK = \angle PAK$, so $P, A, D, K$ are concyclic.

5. Determine Perpendicularity:
- Given $AD \perp PD$ and since $PA = PC$, $PM \perp AC$, so $M$ also lies on the circle $(PADK)$.

6. **Calculate $\angle AKM$:**
$$\angle AKM = \angle APM = 90^\circ - \angle PAC = 90^\circ - \angle ABC$$

7. **Relate $\angle AKC$ and $\angle CKM$:**
$$\angle AKC = 180^\circ - \angle ABC$$
$$\angle AKC = \angle AKM + \angle CKM$$
$$180^\circ - \angle ABC = 90^\circ - \angle ABC + \angle CKM$$
$$180^\circ - \angle ABC = 90^\circ - \angle ABC + \angle CKM$$
$$\angle CKM = 90^\circ$$

Thus, we have proven that $\angle CKM = 90^\circ$.

$\blacksquare$

The final answer is $\boxed{ \angle CKM = 90^\circ }$