4. As shown in Figure 2, $ABCD$ is a square with side length $a$. A circular arc centered at $D$ with radius $DA$ intersects a semicircle with diameter $BC$ at another point $P$. Extending $AP$ intersects $BC$ at point $N$. Then $\frac{BN}{NC}=$ $\qquad$ .
4. $\frac{1}{2}$.

As shown in Figure 8, connect $B P$ and extend it to intersect $A D$ at $E$, connect $C P$ and extend it to intersect $A B$ at $M$, and extend it to intersect the extension of $D A$ at $F$. Clearly, $B E \perp C F$.

Since $M A^{2}=M P \cdot M C$, $M B^{2}=M P \cdot M C$, we have
$M A=M B$.
Also, $\angle A B E=\angle B C M$, thus
Rt $\triangle E A B \cong \mathrm{Rt} \triangle M B C$.
Therefore, $A E=M B=\frac{1}{2} a$.

By Rt $\triangle M A F \backsim \mathrm{Rt} \_C D F$, we know $A F^{\prime}=a$.
Since $\angle E F P=\angle B C P$, we have
Rt $\triangle F P E \backsim \mathrm{Rt} \triangle C P B, \triangle F P A \backsim \triangle C P N$.
Thus, $\frac{N C}{A F}=\frac{C P}{P F}=\frac{B C}{E F}$. Therefore,
$N C=A F \cdot \frac{B C}{E F}=\frac{2}{3} a, B N=a-\frac{2}{3} a=\frac{1}{3} a$.
Hence, $\frac{B N}{N C}=\frac{1}{2}$.