We first prove that the triangles $ADP$ and $CDB$ are similar. Since $ABCD$ is cyclic, $\angle DAP = \angle DCB$; and since $AC$ is tangent to the circle $ADP$, $\angle CBD = \angle CAD = \angle APD$. Consequently, the triangles $ADP$ and $CDB$ are similar.

Let $R$ be the midpoint of $CD$. The points $M$ and $R$ correspond under the above similarity, so $\angle PMA = \angle BRC$.

Let the line $CD$ meet the circle $ABR$ again at $K$ (possibly, $K = R$). Notice that $\angle BAK = \angle BRC = \angle PMA$, so $AK$ is tangent to the circle $AMP$.

Similarly, $BK$ is tangent to the circle $BNQ$ and the conclusion follows.

Alternative solution.
We prove the conclusion under the weaker assumptions that $ABCD$ is merely convex and $AM/DM = CN/BN$.

Let $AB$ and $CD$ meet at $E$; if the two are parallel, then $E$ is their common ideal point. Let $CD$ cross $t_A$ and $t_B$ at $K_A$ and $K_B$, respectively. We will show that $K_A = K_B$, whence the conclusion.

Let $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. The line pencils
($AD, AC, AE, AK_A$) and ($PA, PD, PI, PM$)
are congruent, since $\angle K_A AD = \angle APM$, $\angle CAD = \angle APD$ and $\angle IAE = \angle IPE$. Hence $(D, C; E, K_A) = (AD, AC; AE, AK_A) = (PA, PD; PI, PM) = (A, D; I, M) = DM/AM$. Similarly, $(D, C; E, K_B) = (BD, BC; BE, BK_B) = (QC, QB; QJ, QN) = (C, B; J, N) = BN/CN$.

Finally, recall the assumption $AM/DM = CN/BN$. By the preceding, $(D, C; E, K_A) = (D, C; E, K_B)$, so $K_A = K_B$, as desired. This ends the proof.