Lemma. $AL$ is diameter in $k$. Proof. Denote the circle tangent to $k$ and $BC$ with $\gamma$. There is a homothety centered at at $R$ that sends $\gamma$ to $k$ which also sends the tangent $BC$ to a tangent line to $k$. This line touches $k$ in $S$ so the points $R, P, S$ are collinear. Note that $SI^2=SB^2=SP \cdot SR$ so the triangles $SIP$ and $SIR$ $\angle LRS= \angle IRS= \angle SIP$ whence $AD$ is diameter in $k$. \square Now let $J$ be the symmetric image of $I_a$ under perpendicular bisector of $BC$ and let the altitude from $A$ intersect $k$ second time at $H$. Note that $LS$ is the perpendicular bisector of $II_a$ by the lemma. Thus $\angle JHD \angle I_aLS= \angle ILS= \angle IAR$ and the points $R, H, J$ are collinear. Doing reflection in perpendicular bisector of $BC$ we get the result.