[Solution] Since $P$ is equidistant from $AB$ and $AC$, $AP$ bisects $\angle BAC$.
Also, since $AQ=QP$, we have
$$
\angle APQ=\angle QAP=\angle PAM,
$$

thus $PQ \parallel AM$, which means (2) holds.
Furthermore, by $\text{Rt} \triangle PAM \cong \text{Rt} \triangle PAN$, we have $AM=AN$, which means (1) holds.
However, if $\triangle BMP \cong \triangle QNP$, then
$$
\begin{array}{c}
BP=PQ=AQ, \\
\angle B=\angle QPC=\angle PQC.
\end{array}
$$

Thus, $PC=QC$, so $BC=AC$, which contradicts the fact that $\triangle ABC$ is a scalene triangle. Therefore, (3) does not hold.
Hence, the answer is $(B)$.
17・45 As shown in the figure, in the scalene triangle $A B C$, $A Q=P Q, M P=P N$, $P M \perp A B, P N \perp A C$, then among the following three conclusions:
(1) $A N=A M$;
(2) $Q P / / A M$;
(3) $\triangle B M P \cong \triangle Q N P$.

Which of the following is correct?
(A) (1), (2), (3).
(B) Only (1) and (2).
(C) Only (2) and (3).
(D) Only (1).