Im surprised to see that no one has posted a solution First notice that $MP=2KN \implies MP=AD$ because $\angle{AKD}=90, N \to$ midpoint.And $MP$ is the midsegment so $AC=2MP \implies 2AD=AC$, and this means $\triangle ADC$ is a $30-60-90$ triangle, and this also implies $2DK=DC$ since $\triangle ADC \sim \triangle DKC$. We were given that $MN+DK=DT$ , and since $2DK=DC$ and $2MN=BD$, it follows that $2DT=BC \implies DT=PC$.Let $\angle{ACM}=\alpha \implies \angle{MCP}= 30-\alpha$, and since $MP\parallel AC \to \angle{PMC}=\alpha$, so $\angle{MPC}=150$.Similarly, $\angle{KDC}=60=\angle{BDT}$ and $\angle{ADB}=90$, so $\triangle MPC \cong \triangle ADT$ and we're done!