We will show that $PM$ and $PN$ are the tangents from $P$ to $\omega$; the conclusion then follows at once.

We first prove that $PM$ is the tangent of $\omega$ at $M$. The tangent of $\omega$ at $M$ is parallel to the tangent of $\Gamma$ at $A$, which is in turn parallel to $EF$. Since $MP$ is the parallel through $M$ to $EF$, it follows that $PM$ is the tangent of $\omega$ at $M$.
Next, we show that $PM$ is the tangent of $\gamma$ at $M$. Since the factor $-\frac{1}{2}$ homothety from the barycentre of the triangle $ABC$ exchanges the pairs ($A$, $\Gamma$) and ($M$, $\gamma$), the tangent of $\gamma$ at $M$ is parallel to the tangent of $\Gamma$ at $A$; and since the latter is parallel to $EF$, so is the former. Consequently, $PM$, the parallel through $M$ to $EF$, is the tangent of $\gamma$ at $M$.
By the preceding, $PM$ is the radical axis of $\gamma$ and $\omega$. Since $\omega$ is a homothetic image of $\Gamma$ from $N$, their radical axis is their common tangent at $N$.
Consequently, $PN$ is the other tangent of $\omega$ from $P$ if and only if $P$ is the radical centre of $\Gamma$, $\gamma$ and $\omega$, which is the case if and only if $P$ lies on the radical axis of $\Gamma$ and $\gamma$. Since $P$ lies on $KL$, it is sufficient to show that $KL$ is radical axis of $\Gamma$ and $\gamma$.
Finally, we prove that $KL$ is radical axis of $\Gamma$ and $\gamma$. The powers of $K$ with respect to $\Gamma$ and $\gamma$ are $KA \cdot KB$ and $KD \cdot KE$, respectively. The two are equal since $ABDE$ is cyclic, so $K$ lies on the radical axis of $\Gamma$ and $\gamma$. Similarly, $L$ lies on their radical axis, so $KL$ is the radical axis of the two circles. This completes the solution.