Solution: Let the two roots of the equation be $m_{1}$ and $m_{2}$, then
$$\left\{\begin{array}{l} m_{1}+m_{2}=4-k, \\ m_{1} m_{2}=k . \end{array}\right.$$

Let $P A=x, P B=y, B C=z$, then $x, y, z$ are all positive integers.
By the secant-tangent theorem, we have
$$P A^{2}=P B \cdot P C=P B(P C+3 C),$$

which means
$$x^{2}=y^{2}+y \Rightarrow(x+y)(x-y)=y z \text {. }$$

Eliminating $k$ from (1) and the equation, we get
$$m_{1} m_{2}=4-m_{1}-m_{2} \text {. }$$

Rearranging and factoring, we get
$$\left(m_{1}+1\right)\left(m_{2}+1\right)=5 \text {. }$$

Since the diameter of $\odot O$ is the largest integer root of the equation, it is not difficult to find that the maximum integer root $m=4$. Thus, $z=B C \leqslant 4$.
Since the positive integer $z$ is not a composite number, hence $z=3,2,1$.
When $z=3$, $(x+y)(x-y)=3 y$, we have
$$\left\{\begin{array}{l} x+y=3, \\ x-y=y ; \end{array} \left\lvert\, \begin{array}{l} x+y=y, \\ x-y=3 ; \end{array}\left\{\begin{array}{l} x+y=3 y, \\ x-y=1 . \end{array}\right.\right.\right.$$

The solution that fits the problem is $x=2, y=1$.
When $z=1$ and $z=2$, there are no solutions that fit the problem,
Therefore, $P A=x=2, P B=y=1, P C=y+$ $z=4$.