[Solution] Draw $PE // AB$ intersecting $DC$ at $E$.
$\because \angle BPC > \angle ABP$, then $E$ is on $QC$.
Also, $BQ = QP$, then $\triangle BDQ \cong \triangle PEQ$.
$\therefore \quad S_{\triangle BDQ} = S_{\triangle PEQ} < S_{\triangle PQC}$,
thus
$S_{\triangle ABP} < S_{\triangle ACD}$.
Therefore, the answer is $(A)$.
$17 \cdot 170$ As shown in the figure, in $\triangle A B C$, $P$ is a point on $A C$, take the midpoint $Q$ of $B P$, connect $C Q$ and extend it to intersect $A B$ at $D$. Then the relationship between the area of $\triangle A B P$, $S_{\triangle A B P}$, and the area of $\triangle A C D$, $S_{\triangle A C D}$, is
(A) $S_{\triangle A B P} < S_{\triangle A C D}$.
(B) $S_{\triangle A B P} = S_{\triangle A C D}$.
(C) $S_{\triangle A B P} > S_{\triangle A C D}$.
(D) Cannot be determined.