I’ll just post the exact solution that was in the site since it’s similar to mine: Let the circle centered at $K$ and passing through $B$ meet $AB$ and $BC$ at points $P$ and $Q$ respectively, and let $T$ be the midpoint of arc $ABC$ of the circumcircle. Then $\angle{KPB} = \angle{KBP} = 2\angle{KAP}$, therefore $\angle{KAP} = \angle{PKA}$ and $AP = PK = KB$. Similarly $CQ = QK = KB$. Since $AP = CQ$, $AT = CT$ and $\angle{PAT} = \angle{QCT}$, the triangles $\triangle{TAP}$ and $\triangle{TCQ}$ are congruent i.e. $\angle{TPB} = \angle{TQB}$ and $T$ lies on the circle $BPQ$. Hence the center K of this circle lies on the perpendicular bisector to $BT$ . Furthermore by the assumption $\angle{AKC} = \frac{3\angle{B}}{2}$, i.e. $K$ lies on the corresponding arc. Now let us prove that the constructed point $K$ is in fact the required one. Denote again the common points of the sidelines with the circle centered at $K$ and passing through $B$ by $P$ and $Q$. Since this circle passes through $T$, we obtain that $AP = CQ$. If $AP > PK = KB$ then $\angle{PKA} > \angle{PAK}$, $\angle{KPB} = \angle{KBP} > 2\angle{BAK}$,$\angle{KBC} > 2\angle{KCB}$ and $\angle{AKC} < \frac{3\angle{B}}{2}$ which contradicts to the construction of $K$. Similarly if $AP < PK$ we have $\angle{AKC} > \frac{3\angle{B}}{2} $.

Attachments: