Let $M$ be the midpoint of the segment $CK$. From Menelaus' theorem for the triangle $AKC$ and the line $BN$ we have
$$\frac{\overline{CN}}{\overline{NA}} \cdot \frac{\overline{AB}}{\overline{BK}} \cdot \frac{\overline{KM}}{\overline{MC}} = 1.$$
From this we get $\overline{NA} = 2\overline{NC}$, from which it follows that $\overline{AC} = 3\overline{NC}$. Hence $P_{BNC} = \frac{1}{3}P_{ABC}$. From Menelaus' theorem for the triangle $ABN$ and the line $CU$ we have
$$\frac{\overline{AU}}{\overline{UB}} \cdot \frac{\overline{BL}}{\overline{LN}} \cdot \frac{\overline{NC}}{\overline{CA}} = 1.$$

Therefore we get $\overline{AU} = 3\overline{UB}$. Therefore $U$ is the midpoint of the segment $BK$. It follows that $P_{BUC} = \frac{1}{4}P_{ABC}$. Let $x = P_{CNL}$ and $y = P_{BLU}$. Since $L$ is the midpoint of $BN$, we have $P_{BLC} = x$. Now
$$x + y = P_{BLC} + P_{BLU} = P_{BUC} = \frac{1}{4}P_{ABC},$$
on the other hand we have
$$2x = P_{CNL} + P_{BLC} = P_{BNC} = \frac{1}{3}P_{ABC}.$$
If we divide these two equalities we get
$$\frac{1}{2} + \frac{y}{2x} = \frac{3}{4}, \text{ hence } \frac{y}{x} = \frac{1}{2},$$
from where we get the required statement.