Solution:

Let us first recall that, given two points $A, B$ of the plane and a fixed angle $\alpha$, the locus of points $Q$ such that $\widehat{AQB} = \alpha$ consists of the union of two circular arcs, one for each half-plane determined by the line $AB$.

Let us first identify our locus:

- the two diagonals of the square certainly belong to the locus, indeed if we consider $P \in AC$, then, by symmetry, we have $\widehat{APB} = \widehat{APD}$ and therefore
$$\widehat{APB} + \widehat{CPD} = \widehat{APD} + \widehat{CPD} = 180^{\circ} \text{;}$$

- considering the circumscribed circle of $ABCD$, the arcs $AB$ and $CD$ belong to the locus. Indeed for these points the two angles $\widehat{APB}$ and $\widehat{CPD}$ subtend two arcs of the same length, and since one is acute and one is obtuse, they must be supplementary.

It remains to show that there are no other points of the locus besides these. Notice that, if $P$ is any point of the locus, one of $\widehat{APB}, \widehat{CPD}$ must be at least $90^{\circ}$ and hence $P$ certainly lies inside one of the two circles having $AB$ and $CD$ respectively as diameter. In particular, therefore, $P$ belongs to the strip bounded by the lines $BC$ and $AD$. Let us now distinguish between points interior to the square and points exterior:

- Let $Q$ be interior to the square such that $\widehat{AQB} = \alpha$ and $\widehat{CQD} = 180^{\circ} - \alpha$. However, if $\alpha < 45^{\circ}$ all the points satisfying $\widehat{AQB} = \alpha$ are exterior to the square (the circle will be larger than the one circumscribed about $ABCD$); similarly there cannot be points of the locus interior to the square such that $\widehat{CQD} < 45^{\circ}$. Thus we must certainly have $45^{\circ} \leq \alpha \leq 135^{\circ}$. Let us fix an angle $\alpha$ with this property.

Then there exist at most two points satisfying $\widehat{AQB} = \alpha$ and $\widehat{CQD} = 180^{\circ} - \alpha$; indeed these last two conditions determine (together with being interior to the square), as said at the beginning, two circular arcs, which can intersect in at most two points.

However it is easy to see that the circular arc determined by the condition $\widehat{AQB} = \alpha$, if $45^{\circ} < \alpha < 90^{\circ}$, leaves out the points $C$ and $D$ and has inside it the intersection point of the diagonals of the square $O$, hence it certainly intersects the two diagonals, in two distinct points, which we know belong to the locus. Hence there cannot be any others;

- Let $Q$ be exterior to the square (but still in the strip of the plane determined by the lines $BC$ and $AD$), say in the half-plane determined by $AB$ not containing the square; let us draw the perpendicular to $AB$ through $Q$; it will intersect the arc $AB$ of the circumscribed circle of the square in exactly one point, which we call $P$. It is clear that if $P$ is closer to $AB$ than $Q$ then
$$\widehat{AQB} < \widehat{APB} \quad \widehat{CQB} < \widehat{CPB}$$
and therefore $\widehat{AQB} + \widehat{CQB} < \widehat{APB} + \widehat{CPB} = 180^{\circ}$ (the last equality holds because $P$ belongs to the locus) and similarly we get that if $P$ is farther than $Q$ then $\widehat{APB} + \widehat{CPB} < 180^{\circ}$. Hence necessarily $P = Q$ and therefore there are no other points of the locus exterior to the square that do not lie on the circumscribed circle of $ABCD$.

Alternative solution (for the second part):

Let $P$ be a point of the locus located in the strip bounded by the lines $AB$ and $CD$. Let us call $P'$ the translate of $P$ by the vector $\overrightarrow{BC}$; then, by construction, the quadrilateral $DP'CP$ turns out to be cyclic; but then, since $PP' = CD$, calling $\widehat{CPD} = \alpha$ and $\widehat{CP'D} = 180^{\circ} - \alpha$, two situations can occur:
$$\widehat{PCP'} = \alpha \text{ and } \widehat{PDP'} = 180^{\circ} - \alpha \text{ or vice versa.}$$
In the first case $CP$ and $DP'$ are parallel, hence $AP$ and $CP$ are parallel, that is, $P$ lies on the diagonal $AC$. In the other case, symmetrically, $P$ lies on the diagonal $BD$.

If instead $P$ is outside the strip mentioned above, say in the half-plane determined by the line $AB$ not containing the square, let us consider $P'$ the reflection of $P$ with respect to the perpendicular bisector of the segment $BC$; by construction again $PAP'B$ turns out to be cyclic and the center of the circumscribed circle will lie on the perpendicular bisector of $PP'$, which by construction is the perpendicular bisector of $BC$, and also on the perpendicular bisector of $AB$. Hence the center of this circle is the center of the square and, since $A$ and $B$ are points of the circle, so are $B$ and $C$, by symmetry. Therefore the circle circumscribed about $PAP'B$ is also the circle circumscribed about the square and hence $P$ actually lies on the circumscribed circle of $ABCD$.