Solution:
Let $a, b, c$ be the sidelengths $BC, AC, AB$ respectively, and let $s$ be the semiperimeter of triangle $ABC$ (i.e. let $s = \frac{a + b + c}{2}$). Since $X$ and $Y$ are the points of contact of the incircle we get $AX = AY$. Similarly $BX = BZ$ and $CY = CZ$ where $Z$ is the point of tangency between $\omega$ and side $BC$. Let $p = AX = AY$ and $q = BX = BZ$ and $r = CY + CZ$ as in the diagram.

We have the system of equations:
$$p + q = c$$
$$p + r = b$$
$$q + r = a$$
Adding them together yields $2(p + q + r) = a + b + c$ so therefore $p + q + r = s$. Then we simply get:
$$s - c = (p + q + r) - (p + q) = r.$$
Similarly $p = (s - a)$ and $q = (s - b)$. Hence $BX' = AX = AY = CY' = (s - a)$, and $AX' = (s - b)$.

Now let $\omega_{C}$ be the excircle of triangle $ABC$ opposite vertex $C$. The circle $\omega_{C}$ is tangent to lines $BC$, $AC$ and $AB$ at points $D$, $E$ and $F$ respectively. First we consider equal tangents from $C$ to $\omega_{C}$.
$$CD = BC + BD = a + BF$$
$$CE = CA + AE = b + AF$$
Adding these together gives us $CD + CE = a + b + (AF + BF) = a + b + c$. But since $CD = CE$ (equal tangents) this implies
$$CD = \frac{a + b + c}{2} = s.$$
Therefore $CD = CE = s$. This means $BF = CD - a = (s - a)$ and thus $X'$ and $F$ are the same point.

Now consider the homothety (centred at $C$) that carries $\omega$ to $\omega_{C}$. This homothety sends point $P$ to point $X'$ (since $C, P, X'$ are colinear). It also carries the tangency point $Y$ to $E$. It follows that
$$\frac{CP}{PX'} = \frac{CY}{YE} = \frac{CY}{CE - CY} = \frac{s - c}{s - (s - c)} = \frac{s - c}{c}.$$
i.e. Point $P$ divides segment $CX'$ into the ratio $(s - c):c$. Also we can apply Menelaus' Theorem ($B$, $R$, $Y'$ colinear) to get
$$\frac{CR}{RX'} \times \frac{X'B}{BA} \times \frac{AY'}{Y'C} = 1$$
$$\frac{CR}{RX'} \times \frac{s - a}{c} \times \frac{s - c}{s - a} = 1$$
Hence $\frac{CR}{RX'} = \frac{c}{s - c}$, so point $R$ divides segment $CX'$ into the ratio $c:(s - c)$. Therefore $CP = X'R$ as required.