Prove that $LB = LC$.

Let $\omega$ denote $(MNP)$, and let its circumcentre be $O$.

Note that if $H$ is the orthocentre of $ABC$, then $BN$ and $CP$ concur at $H$ as $MN \parallel AC$, and $MP \parallel AB$ so $BN \perp AC$ and $CP \perp AB$.

Therefore, since $\angle HNM = \angle HPM = 90^{\circ}$, $H$ is the antipodal point of $M$ in $\omega$.

Now, since $L$ is the reflection of $K$ in $\omega$, then $LN$ and $LP$ are tangent to $\omega$. This follows from similar triangles: $OK \times OL = ON^2 \implies \frac{OK}{ON} = \frac{ON}{OL}$, and $\angle LON$ is shared, so $\triangle OKN \sim \triangle ONL \implies \angle ONL = \angle OKN = 90^{\circ}$ as $K$ is the midpoint of $NP$. Therefore, $LN$ is tangent to $\omega$. A similar argument holds for $LP$.

Suppose $P_{\infty}$ is the point at infinity on line $BC$. Then, we have

$$(B, C; M, P_{\infty}) = -1.$$

Projecting these through $H$ onto $\omega$ yields

$$(N, P; M, H') = -1,$$

where $H'$ is the point on $\omega$ such that $HH' \parallel BC$. From this relation, we get $MNH'P$ is a harmonic quadrilateral, which implies $MH'$ is the symmedian of $\triangle MNP$. However, $ML$ is the symmedian of $MNP$ as $L$ is the intersection of the tangents at $N$ and $P$, so we must have $L, H', M$ collinear.

Therefore, since $H'M \perp BC$ (as $\angle HH'M = 90^{\circ}$ and $HH' \parallel BC$), then $LM \perp BC$, so $LB = LC$ as $LM$ is a perpendicular bisector of $BC$.