One of my favourite ones from the entire test . We use Canada 1991 P3 to solve this.

Firstly we begin with a quick lemma. Diagram

$\textbf{LEMMA: }$Let $O$ be a point varying along some fixed circle $\Omega$. Then the circle $\omega$ with center $O$ and a fixed radius $r$ is tangent to some fixed circle. $\textbf{PROOF: }$Let the circle $\Omega$ have center $O'$ and radius $R$. Now let $\Gamma$ denote the circle with center $O'$ and radius $R + r$. So, it's obvious that $\omega$ is tangent to $\Omega$ as $\text{Radius}(\Gamma) = \text{Radius}(\Omega) + \text{Radius}(\omega)$. So our conclusion follows. $\textbf{SOLUTION: }$ Now our problem statement asks us to prove that the $9-$point circle of $\triangle ABC$ is tangent to some fixed circle regardless of the position of $BC$. Let $O$ denote the center of $\odot(ABC)$ and $R$ its radius, $N_9$ its $9-$point center and $H$ its orthocenter. Firstly, note that the radius of the $9-$point circle of $\triangle ABC$ is $= \frac{R}{2}$, which does not depend upon $BC$. So by our Lemma, we are done if we show that $N_9$ varies along some fixed circle. Also, note that by considering a homothety centered at $O$ with scale $2$, $N_9$ maps to $H$ (as $N_9$ is the midpoint of $OH$). Thus it is equivalent to proving that $H$ varies along some fixed circle. So, we move onto proving that the orthocenter of $\triangle ABC$ moves along a fixed circle. So we have from Canada 1991 P3 that the midpoint $M$ of $BC$ varies along a circle. Now let $A'$ denote the $A-$antipode. So by considering a homothety centred at $A'$ with scale $2$, we get that $M$ maps to $H$ and thus we conclude that $H$ too varies along some fixed circle and we are done.

very nice

Sol:- Let $N$ be the nine point center ,$M$ be the midpoint of$AP$ .Let $O'$ be reflection of $O$ across $BC$ .It is known that $N$ is midpoint of $AO'$ so $MN=\frac{OP}{2}$ . Let $R$ be the circumradius of $ABC$ and $MN$ meet nine point circle again at $G$ as in the diag . $MG=\frac{R}{2}-\frac{OP}{2}$. So all the nine point circles are tangent to the circle centered at $M$ with radius $\frac{R}{2}-\frac{OP}{2}$

Attachments:

Denote $\omega$ to be the unit circle. Let $a$ and $p$ be fixed. Then $p=b+c-bc\overline{p}$, i.e. $c=\dfrac{b-p}{b\overline{p}-1}$. Moreover, the midpoint of segment $AP$ is $m=\dfrac{a+p}{2}$, and the $9-$point center is $o_{N}=\dfrac{a+b+c}{2}$. After some manipulation, we obtain that $$\left|o_{N}-m\right|=\left|\dfrac{b+c-p}{2}\right|=\left|\dfrac{(b-p)b\overline{p}}{2(b\overline{p}-1)}\right|=\dfrac{1}{2}\left|bc\overline{p}\right|=\dfrac{1}{2}\left| p \right|$$ which does not depend on $b$. Hence, the $9-$point circle of $\triangle ABC$ is always tangent to some fixed circle centered at $M$, the midpoint of $AP$, as desired.