4. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ have an eccentricity of $e=\frac{\sqrt{3}}{2}$. Given that the point $P\left(0, \frac{3}{2}\right)$ is at a maximum distance of $\frac{7}{4}$ from a point on the ellipse. Then the length of the semi-minor axis $b=(\quad)$.
(A) $\frac{1}{16}$
(B) $\frac{1}{8}$
(C) $\frac{1}{4}$
(D) $\frac{1}{2}$
4. (C).

From $e=\frac{c}{a}=\frac{\sqrt{3}}{2}$, we get $\frac{a^{2}-b^{2}}{a^{2}}=\frac{3}{4}$, which implies $a^{2}=4 b^{2}$.
The equation of the ellipse is $\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1$.
I
Let the distance from a point $(x, y)$ on the ellipse to point $P$ be $d$, then
$$
\begin{aligned}
d^{2} & =x^{2}+\left(y-\frac{3}{2}\right)^{2} \\
& =\left(4 b^{2}-4 y^{2}\right)+\left(y^{2}-3 y+\frac{9}{4}\right) \\
& =-3 y^{2}-3 y+4 b^{2}+\frac{9}{4} \\
& =-3\left(y+\frac{1}{2}\right)^{2}+4 b^{2}+3 .
\end{aligned}
$$

where $y \in[-b, b]$.
(1) If $b \geqslant \frac{1}{2}$, then $y=-\frac{1}{2}$ can make $d^{2}$ (and thus $d$) reach its maximum value. According to the problem, from equation (3) we have
$\left(\frac{7}{4}\right)^{2}=4 b^{2}+3$, which gives $b=\frac{1}{8}$. This contradicts $b \geqslant \frac{1}{2}$.
(2) If $b<\frac{1}{2}$, then by monotonicity, $y=-b$ can make $d^{2}$ (and thus $d$) reach its maximum value. According to the problem, from equation (2) we have $\left(\frac{7}{4}\right)^{2}=b^{2}+3 b+\frac{9}{4}$, which gives $b=\frac{1}{4}$.
The geometric meaning of this problem is that the farthest distance from point $P$ to the ellipse is achieved at the lower vertex of the ellipse's minor axis.
$$
d=\frac{3}{2}+b, b=\frac{7}{4}-\frac{3}{2}=\frac{1}{4} .
$$

If we substitute $d=\frac{7}{4}$ into equation (2) and convert it into a quadratic equation in $y$:
$$
3 y^{2}+3 y-4 b^{2}+\frac{13}{16}=0 \text {, }
$$

and use the discriminant $\Delta=9+4 \times 3\left(4 b^{2}-\frac{13}{16}\right)=0$, we get $b=$ $\frac{1}{8}$. However, in this case, the ellipse $16 x^{2}+64 y^{2}=1$ and the circle $x^{2}+$ $\left(y-\frac{3}{2}\right)^{2}=\left(\frac{7}{4}\right)^{2}$ have no intersection points.