PPOOOPP geometry

Since   \(MR = MK\)   and   \(\overset{\frown}{MK} = 130°\)   ,   \(\overset{\frown}{MR} = 130°\)

\(\overset{\frown}{MK} +\overset{\frown}{KR} +\overset{\frown}{MR} =360°\)

Replace  \(\overset{\frown}{MK}\)  with  130°  and  \(\overset{\frown}{MR}\)  with  130°

\(130°+\overset{\frown}{KR}+130°=360° \\~\\ \overset{\frown}{KR}=360°-130°-130° \\~\\ \overset{\frown}{KR}=100°\)

And..

\(m\angle RPK=\frac{\overset{\frown}{KR}+\overset{\frown}{MQ}}{2}\\~\\ m\angle RPK=\frac{100°+28°}{2} \\~\\ m\angle RPK=64°\)

Draw RK

Since minor arc MK = 130°....then angle KRM  = (1/2)  of this = 65° 

But  MR  = MK...so angle RKM  also  = 65°

And since minor arc MQ = 28°...then angle QKM  = (1/2)  of this = 14°

Therefore, angle RKP  =  m< RKM - m< QKM  = 65 - 14  = 51°

Then, in triangle RKP....angle KRP = angle KRM  = 65°

And angle RKP  = 51°

Therefore, angle  RPK  = 180 - 65 - 51  = 64°

Rats...!!!...hectictar beat me to it!!!

Her method is better, anyway.....no lines had to be drawn and she just dealt with arc measures....much more efficient.....

Here's one other method to solve this  :

Since chord KM = chord RM....then  arc KMR =  130 + 130  = 260°

So....minor arc RK  = 360 - 260  = 100°

And angle RMK is an inscribed angle subtending this arc...so its measure  = 50°

And angle QKM subtends minor arc QM = 28°....so  angle QKM  = 14°

And by  the exterior angle theorem angle RPK  = RMK + QKM   = 50 + 14  = 64° 

[ Note...in essence....this is the same proof as hectictar's   ....!!!  ]

Hey...here's another way!  

∠KRM  =  130° / 2  =  65°

∠RKM  =  130° / 2  =  65°

∠QKM  =  28° / 2  =  14°

∠RKQ  =  65° - 14°  =  51°

And since there are 180° in triangle KPR....     ∠RPK  =  180° - 65° - 51°  =  64°

*edit*

Wait a minute...this isn't any different than the second one!!! Ah..sorry about that!