1. Define the problem setup and notation:
- Let $A$-excircle of $\triangle ABC$ touch $BC$ at $K$ and the extended side $AB$ at $L$.
- Let $B$-excircle touch $BA$ at $M$ and $BC$ at $N$.
- Let $KL$ and $MN$ intersect at $X$.
- We need to show that $CX$ bisects $\angle ACN$.

2. **Use Menelaus' Theorem in $\triangle BMN$ with transversal $KL$:**
- Menelaus' Theorem states that for a transversal intersecting the sides (or their extensions) of a triangle, the product of the ratios of the segments is 1.
- Apply Menelaus' Theorem to $\triangle BMN$ with transversal $KL$:
$$\frac{LB}{LM} \cdot \frac{S_1M}{S_1N} \cdot \frac{KN}{KB} = 1$$
- Substitute the known lengths:
$$\frac{p-c}{a+b} \cdot \frac{S_1M}{S_1N} \cdot \frac{c}{p-c} = 1$$
- Simplify to find:
$$\frac{S_1M}{S_1N} = \frac{a+b}{c} \quad \text{(1)}$$

3. **Use Menelaus' Theorem in $\triangle BMN$ with transversal $VS_2C$:**
- Apply Menelaus' Theorem to $\triangle BMN$ with transversal $VS_2C$:
$$\frac{VM}{VB} \cdot \frac{CB}{CN} \cdot \frac{S_2N}{S_2M} = 1$$
- Substitute the known lengths and expressions:
$$\frac{\frac{(a+b)(p-a)}{a-b}}{\frac{ac}{a-b}} \cdot \frac{a}{p-a} \cdot \frac{S_2N}{S_2M} = 1$$
- Simplify to find:
$$\frac{S_2M}{S_2N} = \frac{a+b}{c} \quad \text{(2)}$$

4. **Conclude that $S_1 \equiv S_2 \equiv S$:**
- From (1) and (2), we have:
$$\frac{S_1M}{S_1N} = \frac{S_2M}{S_2N}$$
- This implies that $S_1$ and $S_2$ are the same point, denoted as $S$.
- Therefore, $KL$ and $CI_b$ are concurrent at $S \in MN$.

5. **Use the lemma to show $CX$ bisects $\angle ACN$:**
- Consider the lemma about the right triangle $XYZ$ and the projections $U$ and $V$.
- Apply this lemma to the right triangle $BI_aI_b$ and the line $d \equiv BA$.
- This implies that $W \in YZ$ and similarly, $CX$ bisects $\angle ACN$.

$$\blacksquare$$