This is the converse of https://artofproblemsolving.com/community/c6h288839p1561572.

Change $\varGamma$ with $C$. Note that since $OT = OP$, by PoP we have that $AP\cdot CP = AT\cdot BT$. Let $D$ be the midpoint of $AT$ and let $E$ be the midpoint of $AP$. It is easy to see that $K,M,E$ are collinear and $\varLambda MD$ are collinear. So we have $\angle KM \varLambda = \angle DME = \angle TAP$. Now we show that $\frac{MK}{M \varLambda} = \frac{AP}{AT}$. Note that $\frac{MK}{M \varLambda} = \frac{BT}{PC}$ and since we know $AP\cdot CP = AT\cdot BT$, we get $\frac{MK}{M \varLambda} = \frac{AP}{AT}$. And so $\triangle APT \sim \triangle MK \varLambda$. Finally $\angle M\varLambda K = \angle ATP = \angle TMK$ as desired. $\square$