4. In the Cartesian coordinate system, $P$ is a moving point on the parabola $y=x^{2}$, point $M\left(0, \frac{1}{4}\right), N\left(0,-\frac{1}{4}\right)$, and a line $l$ is drawn through $N$ parallel to the $x$-axis. Then the positional relationship between the circle $\odot P$ with center $P$ and radius $PM$ and the line $l$ is ( ).
(A) intersecting
(B) tangent
(C) separate
(D) any of the above is possible
4. B.

As shown in Figure 1, draw $PA \perp y$-axis at point $A$, and $PB \perp l$ at point $B$. Let $P\left(a, a^{2}\right)$. Then
$$
\begin{array}{l}
PA=|a|, \\
PB=\left|a^{2}-\left(-\frac{1}{4}\right)\right| \\
=a^{2}+\frac{1}{4} .
\end{array}
$$

In the right triangle $\triangle PAM$, $AM=\left|a^{2}-\frac{1}{4}\right|$.
By the Pythagorean theorem, we get
$$
\begin{array}{l}
PM=\sqrt{AM^{2}+PA^{2}}=\sqrt{\left(a^{2}-\frac{1}{4}\right)^{2}+a^{2}} \\
=\sqrt{\left(a^{2}+\frac{1}{4}\right)^{2}}=a^{2}+\frac{1}{4}=PB .
\end{array}
$$

Therefore, the circle $\odot P$ is tangent to the line $l$.