First, we find angle K
![\begin{gathered} m\angle K+90+56=180 \\ m\angle K=180-90-56=34 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/idypzqz94t0wddk48zmn5o0jxp0uekiu9d.png)
Then, we use the inscribed angle theorem
![\begin{gathered} m\angle K=(1)/(2)\cdot PM \\ PM=2\cdot m\angle K=2\cdot34=68 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nnf91ei6q7hj3cnrtebqm20vg5ovcy0yxy.png)
Hence, the arc PM measures 68°.
Also, the arc MP measures 68° because it's the same arc.
If arc MT is 170°, and MP is 68°, then
![PT=170-68=102](https://img.qammunity.org/2023/formulas/mathematics/college/gs0frwsd1ncd0w2tkwzc953ye13exxptxo.png)
Then, we observe that arc KTP is equal to 180° because it's half the circumference. So,
![\begin{gathered} KT+TP=KTP \\ KT+102=180 \\ KT=180-102=78 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/19gkn70kq4ce6z6jz69x6n4q2k31fnvvvs.png)
Hence, arc KT measures 78°.