Let $X'$ denote the image of $X$. An inversion $\Psi$ around $B$ causes $A',B',C'$ to be collinear, and $K'M'N'$ to be collinear. Then, $\Psi$ also sends the circumcircle of $ACNK$ to the circumcircle $\gamma$ of $A'C'N'K'$. As such, $O'$ lies on the midpoint of the tangents from $\gamma$ to $B$. $M'$ lies on the polar of $B$, hence $\angle M'O'B = 90^{\circ}$. Inverting back implies the result. $\quad \blacksquare$ [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.627272727272732, xmax = 11.245454545454548, ymin = -6.2381818181818245, ymax = 7.652727272727277; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); pen yqqqqq = rgb(0.5019607843137255,0,0); /* draw figures */ draw(circle((-1.1727272727272753,0.16181818181818255), 3.4075754207256947), linewidth(1) + ffxfqq); draw((-3.2454545454545487,5.052727272727277)--(-4.136876511792053,-1.51907132570026), linewidth(1) + red); draw((-4.136876511792053,-1.51907132570026)--(1.1909090909090891,-2.292727272727274), linewidth(1) + red); draw((1.1909090909090891,-2.292727272727274)--(-3.2454545454545487,5.052727272727277), linewidth(1) + red); draw(circle((-0.9928108389069386,1.4008136157456776), 4.290789790551068), linewidth(1) + ffxfqq); draw(circle((-3.4253709792748843,3.8137318387997823), 1.2519902589284826), linewidth(1) + red); draw((-4.662867079350029,3.6237765526121537)--(-1.1727272727272753,0.16181818181818255), linewidth(1) + yqqqqq); draw((-4.662867079350029,3.6237765526121537)--(-3.2454545454545487,5.052727272727277), linewidth(1) + yqqqqq); /* dots and labels */ dot((-1.1727272727272753,0.16181818181818255),dotstyle); label("$O$", (-1.1,0.3436363636363624), NE * labelscalefactor); dot((1.1909090909090891,-2.292727272727274),dotstyle); label("$A$", (1.3,-2.7290909090909126), NE * labelscalefactor); dot((-4.136876511792053,-1.51907132570026),dotstyle); label("$C$", (-4.063636363636367,-1.3290909090909115), NE * labelscalefactor); dot((-3.2454545454545487,5.052727272727277),dotstyle); label("$B$", (-3.172727272727276,5.234545454545457), NE * labelscalefactor); dot((-2.2450261848480517,3.3962803150017655),linewidth(4pt) + dotstyle); label("$K$", (-2.1727272727272755,3.543636363636365), NE * labelscalefactor); dot((-3.5820063289067603,2.57157847640949),linewidth(4pt) + dotstyle); label("$N$", (-3.9909090909090943,2.8709090909090915), NE * labelscalefactor); dot((-4.662867079350029,3.6237765526121537),linewidth(4pt) + dotstyle); label("$M$", (-5.318181818181822,3.6890909090909103), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems See problem 5 from here. The problem is also trivial by Master Miquel.