Angle chasing problem: Find $\angle Q_{2024}Q_{2025}P_{2025}$ in the quadrilateral. 
$ABCD$ is a convex quadrilateral where $BC = CD$, $AC = AD$, $\angle BCD = 96^\circ$ and $\angle ACD = 69^\circ$. Set $P_0 = A, Q_0 = B$ respectively. We inductively define $P_{n+1}$ to be the center of the incircle of $\triangle CDP_n$, and $Q_{n+1}$ to be the center of the incircle of $\triangle CDQ_n$. If $\angle Q_{2024}Q_{2025}P_{2025} - 90^\circ = \frac{2k-1}{2^n}$, compute $k+n$.

This is a question from a national Olympiad. I've been struggling to solve this. Here is my attempt to solve the problem:

In quadrilateral $ABCD$, $\angle CAD=\angle CBD=42^\circ$. So, $A,B,C,D$ are concyclic. So, it can be shown that $C,D,P_n,Q_n$ are concyclic for all $n\geq0$ by induction. So, we have $\angle CQ_nP_n=180^\circ -\angle CDP_n=180^\circ-\frac{69}{2^n}$.
Again, because $\triangle CQ_nQ_{n-1}\cong \triangle CDQ_n$ we have $\angle CQ_nQ_{n-1}=\angle CQ_nD=180^\circ-\angle CDQ_n-\angle DCQ_n$.
Now, $\angle Q_{n-1}Q_{n}P_n - 90^\circ\\=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n-90^\circ\\ =270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ =\angle CDQ_n +\angle DCQ_n+\angle CDP_n-90^\circ\\ =\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}-90^\circ.$
So, I am getting a negative number as answer which is not possible. And I am quite sure that there is nothing wrong in the question. What am I missing and what is the correct solution?

Let us start with the triangle $P_{n-1}CD$ with angles $2x$, $2y$, $2z$ in $P_{n-1},C,D$ respectively. Let $I=P_n$ be its incenter. Then:
$$
\hat P_n=
\hat I
:=
\widehat{CID}=180^\circ-y-z=90^\circ+\frac12(180^\circ-2y-2z)=90^\circ+\frac 12\cdot 2x=90^\circ+\frac 12\hat P_{n-1}\ .
$$
The same relation applies for $\hat Q_n$. This gives
$$
\hat P_n=\hat Q_n=90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}+42^\circ\cdot\frac 1{2^n}\ .
$$
Then we have:
$$
\begin{aligned}
\widehat{Q_{n-1}Q_nP_n}
&=
360^\circ
-\widehat{CQ_nQ_{n-1}}
-\widehat{Q_{n-1}Q_nP_n}
\\
&=
360^\circ
-\widehat{CQ_nD}
-\left(180^\circ-\widehat{CDP_n}\right)
\\
&=
180^\circ
-\widehat{CQ_nD}
+\frac 1{2^n}\widehat{CDA}
\\
&=
180^\circ
-90^\circ\cdot\frac{1-\frac 1{2^n}}{1-\frac 12}-42^\circ\cdot\frac 1{2^n}
+69^\circ\cdot\frac 1{2^n}
\\
&=
180^\circ
-180^\circ\cdot\left(1-\frac 1{2^n}\right)
+27^\circ\cdot\frac 1{2^n}
\\
&=
\left(
180^\circ
+27^\circ
\right)
\cdot\frac 1{2^n}
\ .
\end{aligned}
$$

Where is the error in the following lines?
$$
\begin{aligned}
\angle Q_{n-1}Q_{n}P_n\color{blue}{ - 90^\circ}
&=360^\circ -\angle CQ_nQ_{n-1}-\angle CQ_nP_n\color{blue}{ - 90^\circ}
\\ 
&=270^\circ-(180^\circ-\angle CDQ_n-\angle DCQ_n)-(180^\circ-\angle CDP_n)\\ &=\angle CDQ_n +\angle DCQ_n+\angle CDP_n\color{blue}{ - 90^\circ}\\ 
&=\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ}
\end{aligned}
$$
No error, $\angle Q_{n-1}Q_{n}P_n \color{blue}{ - 90^\circ}=
\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}\color{blue}{ - 90^\circ}$
gives the same
$\angle Q_{n-1}Q_{n}P_n =
\frac{42^\circ}{2^n}+\frac{96^\circ}{2^n}+\frac{69^\circ}{2^n}$.
The angle is so small, since the arcs $CP_{n-1}Q_{n-1}D$ and $CP_nQ_nD$ are hardly distinguishable in some computer picture, they are plotted slightly over the segment $CD$, so we expect an angle which is either almost $0^\circ$ or $180^\circ$.
