B

$\because AP$、$CP$分别是$\triangle ABC$的外角$\angle ACE$和外角$\angle CAF$的平分线，
$\therefore ∠PAC=\frac{1}{2}∠CAF$，$∠ACP=\frac{1}{2}∠ACE$，
$\because \angle P=70^{\circ}$，
$\therefore \angle PAC+\angle PCA=180^{\circ}-70^{\circ}=110^{\circ}$，
$\therefore \angle CAF+\angle ACE=2\left(\angle PAC+\angle PCA\right)=220^{\circ}$，
$\because \angle FAC+\angle BAC=180^{\circ}$，$\angle ECA+\angle ACB=180^{\circ}$，
$\therefore \angle BAC+\angle BCA=180^{\circ}+180^{\circ}-\left(\angle FAC+\angle ECA\right)=140^{\circ}$，
$\therefore \angle B=180^{\circ}-\left(\angle BAC+\angle BCA\right)=40^{\circ}$，故$B$正确.
故选：$B$.