B. As shown in Figure 7, the acute triangle $\triangle ABC$ is inscribed in $\odot O, AB > AC, BC$ is extended to a point $P$ and a point $Q$ on the segment $BC$ such that $PA$ is tangent to $\odot O$, and $\angle POQ + \angle BAC = 90^{\circ}$. Let $\frac{PA}{PO} = t$, express $\frac{PQ}{PC}$ in terms of $t$.
Figure 7
B. Let the radius of $\odot O$ be $r$, and connect $O B, O C$.

By the power of a point theorem, we have
$$
\begin{array}{l}
P C \cdot P B = P O^{2} - r^{2} = P A^{2}, \\
Q C \cdot Q B = r^{2} - Q O^{2}. \\
\text{Since } \angle O C Q = 90^{\circ} - \angle B A C = \angle P O Q \\
\Rightarrow \triangle O Q C \sim \triangle P Q O \\
\Rightarrow Q O^{2} = Q C \cdot Q P.
\end{array}
$$

From equations (2) and (3), we get
$$
\begin{array}{l}
Q C \cdot P B = Q C \cdot Q B + Q C \cdot Q P \\
= r^{2} - Q O^{2} + Q O^{2} = r^{2} = P O^{2} - P A^{2}.
\end{array}
$$

Combining equation (1), we have
$$
\frac{Q C}{P C} = \frac{Q C \cdot P B}{P C \cdot P B} = \frac{P O^{2} - P A^{2}}{P A^{2}}.
$$

Noting that, $\frac{P O}{P A} = \frac{1}{t}$.
$$
\begin{aligned}
& \text{Thus } \frac{P Q}{P C} = \frac{Q C + P C}{P C} = 1 + \frac{P O^{2} - P A^{2}}{P A^{2}} \\
= & \frac{P O^{2}}{P A^{2}} = \left(\frac{P Q}{P A}\right)^{2} = \frac{1}{t^{2}}.
\end{aligned}
$$