The center of a circle passing through points $M$ and $N$ is on the perpendicular bisector $y = 3 - x$ of $MN$. Denote the center by $S(a, 3-a)$, then the equation of the circle $S$ is
$$(x-a)^2 + (y-3+a)^2 = 2(1+a^2).$$
Since for a chord with a fixed length, the angle at the circumference subtended by the corresponding arc will become larger as the radius of the circle becomes smaller. When $\angle MPN$ reaches its maximum value, the circle $S$ through the three points $M$, $N$ and $P$ will be tangent to the $x$-axis at $P$, which means the value $a$ in the equation of $S$ has to satisfy the condition $2(1+a^2) = (a-3)^2$. Solve the above equation we have $a = 1$ or $a = -7$. Thus the points of contact are $P(1, 0)$ and $P'(-7, 0)$ respectively.
But the radius of the circle through points $M$, $N$, and $P'$ is larger than that of the circle through points $M$, $N$ and $P$. Therefore $\angle MPN > \angle MP'N$. Thus $P(1, 0)$ is the point we want to find, and the $x$-axis of point $P$ is 1.