We use barycentric coordinates with $\triangle{ABC}$ as the reference triangle. We have,
$a^2 + c^2 = b^2.$ Since $Q$ is a arc midpoint, it follows that $AQ$ passes through the incenter. In other words, we parameterize $Q = (t : b : c).$ Plugging this into the circumcircle equation gives,
$$a^2bc + b^2ct + c^2bt = 0 \rightarrow t = \left(-\dfrac{a^2}{b+c} : b : c \right) .$$In a similar fashion we have
$$P = \left(a : b : -\dfrac{c^2}{a+b} \right).$$Trivially,
$$K = (0 : b : c) \quad\,\,\text{and}\quad\,\, L=(a : b : 0)$$Also,
$$M = (1 : 1 : 0) \quad\,\,\text{and}\quad\,\, N = (0 : 1 : 1).$$Therefore, the equation for line $KL$ is
$$
\begin{vmatrix}
0 & b & c \\
a & b & 0 \\
x & y & z
\end{vmatrix}
= x(-bc)+y(ac)+z(-ab)=0.
$$The equation for line $MQ$ is
$$
\begin{vmatrix}
1 & 1 & 0 \\
-\dfrac{a^2}{b+c} & b & c \\
x & y & z
\end{vmatrix}
= x(c) + y(-c) + z\left(\dfrac{a^2}{b+c} + b \right) =0.
$$The equation for line $NP$ is
$$
\begin{vmatrix}
0 & 1 & 1 \\
a & b & -\dfrac{c^2}{a+b} \\
x & y & z
\end{vmatrix}
= x \left(-\dfrac{c^2}{a+b}-b \right) + y(a)+z(-a)= 0.
$$By the concurrency lemma, it suffices to show that
$$
\begin{vmatrix}
-bc & ac & -ab \\
c & -c & \dfrac{a^2}{b+c} + b \\
-\dfrac{c^2}{a+b} - b & a & -a
\end{vmatrix}
=0.
$$However, the determinant in question turns out to be
$$ab^2c(b^2 - a^2 - c^2) = 0 $$by the Pythagorean Theorem.