Sides $A B$ and $A C$ of $\triangle A B C$ are produced to points $D$ and $E$, respectively. The bisectors of $\angle C B D$ and $\angle B C E$ meet at $P$. If $\angle A=78^{\circ}$, then the measure of $\angle P$ is:

51°

Given \(\angle\)A = \({78}^\circ\)

If the bisectors of \(\angle\)CBD and \(\angle\)BCE meet at P

\(\angle\)P = \({90}^\circ\) - \(\angle\)A/2

⇒ \(\angle\)P = \({90}^\circ\) - \(\angle\)78/2

⇒ \(\angle\)P = \({90}^\circ\) - \({39}^\circ\) = \({51}^\circ\)

Therefore, \(\angle\)P is \({51}^\circ\)