The answer is C

\(\Delta PQS\) has PQ = PS,so \(\Delta PQS\) isosceles at P

\(\Rightarrow\widehat{Q}=\widehat{PSQ}\) but \(\widehat{QPS}+\widehat{Q}+\widehat{PSQ}=180^0\)

\(\Rightarrow2\widehat{PSQ}+12^0=180^0\Rightarrow\widehat{PSQ}=84^0\)

\(\Delta PSR\) has SP = SR,so \(\Delta PSR\) isosceles at S

\(\Rightarrow\widehat{SPR}=\widehat{R}\)

\(\widehat{PSQ}\) is the exterior angle of \(\Delta PSR\Rightarrow\widehat{PSQ}=\widehat{SPR}+\widehat{R}\)

\(\Rightarrow2\widehat{SPR}=84^0\Rightarrow\widehat{SPR}=42^0\Rightarrow\widehat{QPR}=12^0+42^0=54^0\)

Hence,the answer is C

\(\Delta PQS\) has: \(PQ=PS\)

\(\Rightarrow\)​\(\Delta PQS\) is an ​ isosceles triangle.

\(\Rightarrow\)\(\widehat{PSQ}=\frac{180^0-\widehat{QPS}}{2}\)

\(\Rightarrow\)\(\widehat{PSQ}=84^0\)

\(\Rightarrow\)\(\widehat{PSR}=180^0-84^0=96^0\)

\(\Delta PSR\) has: \(SP=SR\)

\(\Rightarrow\)\(\Delta PSR\) is an isosceles triangle

\(\Rightarrow\)\(\widehat{SPR}=\frac{180^0-\widehat{PSR}}{2}\)

\(\Rightarrow\)\(\widehat{SPR}=42^0\)

\(\Rightarrow\)\(\widehat{QPR}=12^0+42^0=54^0\)

Answer: C.