Let $X,Y$ and $Z$ be midpoints of sides $KN,KM$ and $ML$ respectively. Note that by Midpoint thm, $MN=2PQ$ and $KL=2QT$. Claim : $NK\geq 2AP$ and $ML\geq 2CT$. Proof : Let FSOC $NK<2AP$. So $NQ,QK<AP$. Hence we obtain $\angle QAN<\angle ANQ, \angle QAK<\angle AKQ$. Thus $2\angle KAN=2(\angle QAN+\angle QAK)<\angle QAN+\angle QAK+\angle AKQ+\angle ANQ=180^{\circ}$ and thus $\angle A<90^{\circ}$ which is a contradiction as angle $A$ isn't acute. So we obtain $NK\geq 2AP$. Similarly $ML\geq 2CT$ and our claim is proven. We finally obtain $KL+LM+MN+NK\geq 2(AP+PQ+QT+TC)\geq 2AC$ which gives us that perimeter of $KLMN$ is greater than or equal to $2AC$ as desired.