We do inversion at $C$ with some radius. Let $X'$ denote image of $X$ Then $(PAC) \leftrightarrow P'A'$ and $(PBC) \leftrightarrow P'B'$ Let $m_1$ and $m_2$ be common tangent to $(PAC)$ and $(PBC)$. As they are tangent to two circle we get they convert into circle tangent line $P'A'$ and $P'B'$ and pass through $C$. Let circles be $w_1$ and $w_2$.($m_1 \leftrightarrow w_1$ and $m_2 \leftrightarrow w_2$) Observe from $Q=m_1\cap m_2$ we get $Q'$ is intersection of $w_1$ and $w_2$ other then $C$. Now $w_1$ and $w_2$ tangent to $P'A'$ and $P'B'$ hence center of two circle should lie on angle bisector of $\angle A'P'B'$.Let angle bisector be line $h$. Let $(PAB) \leftrightarrow (P'A'B')$ (Refer $(P'A'B')$ as $\Gamma$) and $W=h \cap \Gamma$ other then $P'$.\newline \newline Note as $h$ is angle bisector, $W$ will be midpoint of arc $A'B'$.And as $A'$ and $B'$ are fixed we get $W$ is fixed. $h$ pass though center of both circle hence its should perpendicular bisect $Q'C$ (as they are intersection of $w_1$ and $w_2$). Hence $\Box P'CWQ'$ is kite with $P'C=P'Q$ and $WC=WQ'$. Now as $P$ move on circle, $P'$ also move on $\Gamma$ and so $h$ and so $Q'$.But as $WC=WQ'$ and $WC$ is fixed length. Hence we get $Q'$ move on circular path $k_1$. Also as $WC$ is radius of that circular path $C$ also lie on $k_1$. Circle remain till $P'$ lie between major arc of $A'B'$.. When $P'$ enter in minor arc of $A'B'$ we get $h$ pass thought midpoint of arc $A'B'$ not contain $P'$. Let it be $Y$. Then again $YC=YQ'$ and hence $Q'$ move on circular path $k_2$, with radius $YC$ center at $Y$(which also pass through $C$). Now circle in inversion convert into line as they pass though $C$. Hence $k_1$ and $k_2$ convert into lines. As $k_1$ and $k_2$ was locus of $Q'$ we get $Q$ lie on two line which are image of $k_1$ and $k_2$.$\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.76964995836879cm); 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 = -13.940165964997371, xmax = 18.829483993371415, ymin = -10.885499950029526, ymax = 6.237030419213004; /* image dimensions */ /* draw figures */ draw(circle((0.,0.), 6.), linewidth(0.4) + red); draw((xmin, -13.818910280058669*xmin-55.39025375368372)--(xmax, -13.818910280058669*xmax-55.39025375368372), linewidth(0.4)); /* line */ draw((xmin, -1.073653703835097*xmin-0.45395423647996425)--(xmax, -1.073653703835097*xmax-0.45395423647996425), linewidth(0.4)); /* line */ draw((xmin, -2.294105516521462*xmin-5.7145079120434366)--(xmax, -2.294105516521462*xmax-5.7145079120434366), linewidth(0.4)); /* line */ draw(circle((0.12388834027765272,-5.998720836907086), 4.806888667502303), linewidth(0.4)); /* dots and labels */ dot((-3.6644935696393914,-4.750945892985054),dotstyle); label("$A'$", (-3.476397406015869,-4.556764235320835), NE * labelscalefactor); dot((3.857522685583134,-4.595597755484207),dotstyle); label("$B'$", (3.745350096657821,-5.177989826948682), NE * labelscalefactor); dot((-4.310332961023954,4.173850711885905),dotstyle); label("$P'$", (-4.7382618890099275,3.383275357672582), NE * labelscalefactor); dot((-0.4196452328159629,-1.222660753880266),dotstyle); label("$C$", (-0.3508561481382771,-1.0235436829374565), NE * labelscalefactor); dot((-3.005088396432255,-2.3496547587317007),dotstyle); label("$Q'$", (-3.321091008108908,-1.9942086698559673), NE * labelscalefactor); dot((0.12388834027765272,-5.998720836907086),linewidth(4.pt) + dotstyle); label("$W$", (0.19271624453608666,-5.6633223204079375), NE * labelscalefactor); dot((-0.12388834027765451,5.9987208369070855),linewidth(4.pt) + dotstyle); label("$Y$", (0.2121295442744568,5.596391527846786), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Make an inversion at point $C$ and note that the points $Q'$ and $C$ are symmetric with respect to one of the bisectors of the angles $A'P'B'$, from where point $Q'$ lies on two circles passing through point $C$, the centers of which are the midpoints of the arcs $A'B'$ of the circle, from which the required follows.

Cool problem! starchan wrote: A point $P$ moves along a circle $\Omega$. Let $A$ and $B$ be two fixed points of $\Omega$, and $C$ be an arbitrary point inside $\Omega$. The common external tangents to the circumcircles of triangles $APC$ and $BCP$ meet at point $Q$. Prove that all points $Q$ lie on two fixed lines. Invert at $C$ to get that $Q$ is the intersection of two circles passing through $C$ tangent to lines $PA$ and $PB$, where $A, B$ are fixed points and $P$ is a moving point on a fixed circle passing through them. These two circles are symmetric in the internal bisector of angle $APB$, hence $Q$ is simply the reflection of $C$ in the internal bisector of angle $APB$. Let $M, N$ be the two midpoints of arcs $AB$, then $Q$ lies on the union of the two circles with centres at $M$ and $N$ respectively passing through $C$. Invert back to get that $Q$ lies on two fixed lines as desired.