Let $\Gamma_1,\Gamma_2,\Gamma_3$ be the circle $(BEZG)$,$(BNL)$ and $(LZK)$ respectively. $\underline{CLAIM}$:$HZ$ is tangent to $\Gamma_1$ $\underline{Proof.}$Let $X$ be the point such that $BACX$ is an isoceles trapezoid, now we have $$-1=(A,C;M,\infty_{AC})=(E,G;Z,X)$$Therefore $EH$ is the $E-symmedian$ of $\triangle EZX$, together with the fact that $MH$ is the perpendicular bisector of $ZX$, hence $H$ is the intersection of the tangent at $Z$ and $X$ at $\Gamma_1$, this completes the proof. Now we have $\angle HZB=\angle HZM=90^{\circ}$, hence $BNZK$ is cyclic, by power of a point $$HB\times HN=HK\times HZ$$Hence $H$ lies on the radical axis of $\Gamma_2$ and $\Gamma_3$, that is $LP$, we conclude that $BZ$, $KN$ and $HP$ concurrentn at $L$.

parmenides51 wrote: A triangle $ABC$ is given. Let $M$ be the midpoint of the side $AC$ of the triangle and $Z$ the image of point $B$ along the line $BM$. The circle with center $M$ and radius $MB$ intersects the lines $BA$ and $BC$ at the points $E$ and $G$ respectively. Let $H$ be the point of intersection of $EG$ with the line $AC$, and $K$ the point of intersection of $HZ$ with the line $EB$. The perpendicular from point $K$ to the line $BH$ intersects the lines $BZ$ and $BH$ at the points $L$ and $N$, respectively. If $P$ is the second point of intersection of the circumscribed circles of the triangles $KZL$ and $BLN$, prove that, the lines $BZ, KN$ and $HP$ intersect at a common point. What do you mean by "Image of $B$ along the line $BM$"? $B$ is already on line $BM$ so what do you mean by its image?

I guess it's reflection of $B$ on $M$, am I correct?

yes, it means that