In a cyclic quadrilateral $ABCD$,diagonals $AC$ and $BD$ intersect at $P$. If $\angle BAC = 52^{\circ}$ and $\angle ADB = 78^{\circ}$,then find $\angle ABC$. (in $^{\circ}$)

  • A
    $90$
  • B
    $60$
  • C
    $45$
  • D
    $50$

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