Solution:

The perpendicular bisector of the segment $AB$ and the bisector of $\text{ACB}$ meet at the midpoint $D$ of the arc $\text{AB}$ of the circumcircle of $\triangle ABC$ which does not contain $C$.

Next we use the standard notation for the elements of $\triangle ABC$. Since $AM = BN = CP = CQ = p-c$, the condition of the problem is equivalent to the equality $DM = DP$. The Cosine theorem gives
$$\begin{aligned} DP^{2} & = DC^{2} + CP^{2} - 2 DC \cdot CP \cos \frac{\gamma}{2} \\ DM^{2} & = DA^{2} + AM^{2} + 2 DA \cdot AM \cos \frac{\gamma}{2} \end{aligned}$$
Subtracting these equalities, we get
$$DC - DA = 2(p-c) \cos \frac{\gamma}{2}$$
On the other hand, we have $DC = \frac{(a+b) DA}{c}$ by Ptolemy's theorem. Since $DA = \frac{c}{2 \cos \frac{\gamma}{2}}$, we get $DC = \frac{a+b}{2 \cos \frac{\gamma}{2}}$. Now (1) implies that $\cos^{2} \frac{\gamma}{2} = \frac{1}{2}$, i.e. $\gamma = 90^{\circ}$.

Remark. The solution above shows that the points $M, N, P$ and $Q$ are concyclic if and only if $AC = BC$ or $\text{ACB = 90}$.