4. As shown in Figure 3, let $\odot O$ be the circumcircle of the unit square $ABCD$, and point $P$ lies on the minor arc $AB$ (excluding $A$ and $B$). $DP$ intersects $AO$ at point $Q$. Then the range of $k=\frac{S_{\triangle AQD}}{S_{\triangle OPC}}$ is
4. $[2 \sqrt{2}-2,1)$.

Let the radius of $\odot O$ be $r$. By the Pythagorean theorem, it is easy to get $r=\frac{1}{\sqrt{2}}$.
Let $\angle O P C=\angle O C P=\angle A C P=\angle A D P=\theta\left(0^{\circ}<\theta<45^{\circ}\right)$. Draw $Q T \perp A D$, with $T$ as the foot of the perpendicular, and let $Q T=h$. Then
$$
1=A D=D T+T A=h \cot \theta+h \cot 45^{\circ},
$$

which means $h=\frac{1}{1+\cot \theta}$.
By the area formula, we get
$$
\begin{array}{l}
\frac{S_{\triangle A Q D}}{S_{\triangle O F C}}=\frac{\frac{1}{2} \times 1 \times h}{\frac{1}{2} r^{2} \sin \left(180^{\circ}-2 \theta\right)} \\
=\frac{1}{1+\cot \theta} \cdot \frac{2}{\sin 2 \theta}=\frac{2}{2 \sin \theta \cdot \cos \theta+2 \cos ^{2} \theta} . \\
=\frac{2}{\sin 2 \theta+\cos 2 \theta+1}=\frac{2}{1+\sqrt{2} \sin \left(45^{\circ}+2 \theta\right)} .
\end{array}
$$

Since $45^{\circ}<45^{\circ}+2 \theta<45^{\circ}+90^{\circ}$, we have
$$
\frac{1}{\sqrt{2}}<\sin \left(45^{\circ}+2 \theta\right) \leqslant 1 \text {. }
$$

Therefore, $\frac{2}{1+\sqrt{2}} \leqslant \frac{S_{\triangle A Q D}}{S_{\triangle O P C}}<1$.