Reflect $A$ across $D$ to a new point $Q$. Then $\triangle QDE\cong \triangle ECB$ (sas).
Since $$\angle QEC = \angle DEB $$ we see that quadrilateral $ACEQ$ is cyclic, so $$\angle QCE =\angle QAE = \angle DAE = \angle DNE$$ so $QC||DN$. Now since $D$ halves $AQ$ the line $DI$ is a middle line for triangle $QAC$ and thus it halves $AC$.
Ignoring vertex $A$, this becomes a problem on the inscriptable $\square BCNM$. Let $L$ be the midpoint of $\overline{BC}$. Also, let $M'$ (instead of $P$), $N'$ (instead of $Q$), $H'$, $K'$ be the projections of $M$, $N$, $H$, $K$ onto $\overline{BC}$. Let the tangent segments from $B$ and $C$ to the incircle have length $d$; let the tangent segments from $M$ and $N$ have length $m$ and $n$. Finally, define $m' := |MM'|$, $n':=|NN'|$, $m'':=|M'L|$, $n'':=|N'L|$. Without loss of generality, $m\leq n$ so that $m'\leq n'$.
Certainly, if $m=n$, then $\overleftrightarrow{HK}$ meets $\overline{BC}$ at $L$. For $m \neq n$, we'll prove that $L$ is on $\overleftrightarrow{HK}$ by showing $\triangle HH'L\sim \triangle KK'L$ via
$$|HH'||K'L|=|KK'||H'L| \tag{$\star$}$$
Parallelism and proportionality rules tell us that
$$\frac{|M'H'|}{|M'N'|}=\frac{|MH|}{|MN|}=\frac{m}{m+n} \qquad |HH'|=m'+\frac{m}{m+n}(n'-m')=\frac{m'n+mn'}{m+n} \tag{1}$$
The Crossed Ladders Theorem tells us that
$$\frac{1}{|KK'|}=\frac{1}{m'}+\frac{1}{n'} \quad\to\quad |KK'| = \frac{m'n'}{m'+n'} \tag{2}$$
(and, in fact, $K$ is the midpoint of the extension of $\overline{KK'}$ that meets $\overline{MN}$), whereupon some proportional thinking then yields $|M'K'|:|K'N'|=m':n'$, so that we have
$$\frac{|M'K'|}{|M'N'|}=\frac{m'}{m'+n'} \tag{3}$$
Therefore,
$$\begin{align}
|H'L|&=|M'L|-|M'H'| = m'' - \,\frac{m}{m+n}(m''+n'') \;= \frac{m''n-mn''}{m+n} \\[6pt]
|K'L|&=|M'L|-|M'K'| = m'' - \frac{m'}{m'+n'}(m''+n'')=\frac{m''n'-m'n''}{m'+n'}
\end{align} \tag{4}$$
Substituting in $(\star)$, and clearing denominators, we need only verify that
$$(m'n+mn')(m''n'-m'n'') = m'n'(m''n-mn'') \tag{5} $$
That is,
$$\frac{m}{n}\cdot\frac{m''}{n''} = \left(\frac{m'}{n'}\right)^2 \tag{6}$$
It seems like there's a geometric mean argument to be made, but I'm not seeing it. So, writing $\theta$ for the common angle at $B$ and $C$, we have
$$\frac{m}{n}\cdot\frac{d-(m+d)\cos\theta}{d-(n+d)\cos\theta} = \left(\frac{m+d}{n+d}\right)^2 \quad\to\quad (d+m)(d+n)\cos\theta = d^2 - m n \tag{7}$$
This same relation results (for $\theta \neq 0$) from the observation that there's a right triangle with hypotenuse $|MN|$ and legs $|m'-n'|$ and $m''+n''$.
$$(m+n)^2 = (m'-n')^2 + (m''+n'')^2 \qquad\to\qquad (7) \tag{8}$$
This equality establishes $(\star)$ and completes the proof. $\square$
I believe there's a cleaner way to link $(6)$ and $(8)$ (or to demonstrate $(6)$ some other way) without having to show equality through $(7)$. Again, I'm not seeing it. Perhaps I'll return to this question.
Best Answer
This was not as easy as I thought.
Draw the second tangent (other than $CP$) from $C$ to the circle with centre $O$ and call it $CJ$. Define $E'$ as the intersection of $AB$ and $PJ$.
Now, since $PBJA$ is harmonic, the tangents from $A$ and $B$, and $PJ = PE$ must concur. It is not hard to see that they concur on $OM$.