We first prove that triangles $ADQ$ and $CDB$ are similar. Since $ABCD$ is cyclic, we have $\angle DAQ = \angle DCB$. By the tangency of $AC$ to the circle $AQD$ we also have $\angle CBD = \angle CAD = \angle AQD$. The claimed similarity is proven.

Let $R$ be the midpoint of $CD$. Points $N$ and $R$ correspond in the proven similarity, and so $\angle QNA = \angle BRC$.

Let $K$ be the second common point of line $CD$ with circle $ABR$ (i.e., if $CD$ intersects circle $ABR$, then $K \neq R$ is the other intersection; otherwise, if $CD$ is tangent to $CD$, then $K = R$). In both cases, we have $\angle BAK = \angle BRC = \angle QNA$; that indicates that $AK$ is tangent to circle $ANQ$. It can be showed analogously that $BK$ is tangent to circle $BMP$.

We present a second solution, without using the condition that $ABCD$ is cyclic. Again, $M$ and $N$ can be any points on lines $BC$ and $AD$ such that $BM : MC = DN : NA$.

Let $AB$ and $CD$ meet at $T$ (if $AB \parallel CD$ then $T$ is their common ideal point). Let $CD$ meet the tangent to the circle $ANQ$ at $A$, and the tangent to the circle $BMP$ at $B$ at points $K_1$ and $K_2$, respectively.

Let $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. Notice that the pencils ($AD$, $AC$, $AT$, $AK_1$) and ($QA$, $QD$, $QI$, $QN$) of lines are congruent, because $\angle K_1AD = \angle AQ N$, $\angle CAD = \angle AQD$ and $\angle IAT = \angle IQT$. Hence,

$$(D, C ; T, K_1) = (AD, AC ; AT, AK_1) = (QA, QD ; QI, QN) = (A, D ; I, N) = \frac{DN}{NA}.$$

It can be obtained analogously that

$$(D, C ; T, K_2) = (BD, BC ; BT, BK_2) = (PC, PB ; PJ, PM) = (C, B ; I, N) = \frac{BM}{MC}.$$

From $BM : MC = DN : DA$ we get $(D, C ; T, K_1) = (D, C ; T, K_2)$ and hence $K_1 = K_2$.