(1) As shown in Fig. 1, by the Ptolemy inequality we have
$$PA \times BC + PC \times AB \ge PB \times AC.$$
Therefore,
$$\begin{aligned} f(P) &= PA \times BC + PC \times AB + PD \times CA \\ &\ge PB \times CA + PD \times CA \\ &= (PB + PD) \times CA. \end{aligned}$$
The equality holds if and only if $P$, $A$, $B$, $C$ lie on $\odot O$ and $P$ on $\widehat{AC}$. Furthermore, $PB + PD \ge BD$, and the equality holds if and only if $P$ lies on line $BD$. Combining the results above, we have
$$f(P)_{\min} = BD \times CA.$$

(2) Denote $\angle ECB = \alpha$. Then $\angle ECA = 2\alpha$. By the sine theorem we have
$$\frac{AE}{AB} = \frac{\sin 2\alpha}{\sin 3\alpha} = \frac{\sqrt{3}}{2}.$$
That is to say, $\sqrt{3} \sin 3\alpha = 2\sin 2\alpha$. By using trigonometric identities,
$$\sqrt{3}(3\sin \alpha - 4\sin^3 \alpha) = 4\sin \alpha \cos \alpha.$$
By simplification,
$$3\sqrt{3} - 4\sqrt{3}(1 - \cos^2\alpha) - 4\cos\alpha = 0.$$
$$\text{That is to say, } 4\sqrt{3}\cos^2\alpha - 4\cos\alpha - \sqrt{3} = 0.$$
$$\text{The solutions are } \cos \alpha = \frac{\sqrt{3}}{2} \text{ and } \cos \alpha = -\frac{1}{2\sqrt{3}} \text{ (discarded).}$$
$$\text{Therefore, } \alpha = 30^\circ \text{ and } \angle ECA = 60^\circ.$$
On the other hand,
$$\frac{BC}{EC} = \sqrt{3} - 1 = \frac{\sin(\angle EAC - 30^\circ)}{\sin \angle EAC}.$$
That is to say,
$$\frac{\sqrt{3}}{2}\sin\angle EAC - \frac{1}{2}\cos\angle EAC = (\sqrt{3} - 1)\sin\angle EAC.$$
Then
$$\frac{2 - \sqrt{3}}{2} \sin \angle EAC = \frac{1}{2} \cos \angle EAC.$$
Therefore,
$$\tan \angle EAC = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}.$$
$$\text{We obtain } \angle EAC = 75^\circ \text{ and } \angle AEC = 45^\circ.$$
Since $\triangle ADC$ is an isosceles triangle and $AC = \sqrt{2}$, we have $CD = 1$. Furthermore, $\triangle ABC$ is an isosceles triangle, so $BC = \sqrt{2}$ and $AB = 2$. Then
$$BD^2 = AB^2 + AD^2 = 4 + 1 = 5.$$
$$\text{We have } BD = \sqrt{5}. \text{ Therefore,}$$
$$$$
f(P)_{\min} = BD \times CA = \sqrt{5} \times \sqrt{2} = \sqrt{10}.

(1) As shown in Fig. 2, the line $BD$ intercepts the circumscribed circle $O$ of $\triangle ABC$ at point $P_0$, which lies on $BD$ because $D$ is outside of $\odot O$. Through $A$, $C$, $D$ draw lines perpendicular to $P_0A$, $P_0C$, $P_0D$, respectively, and they constitute $\triangle A_1B_1C_1$ by intercepting with each other. It is easy to see that $P_0$ lies in $\triangle ABC$, as well as in $\triangle A_1B_1C_1$. Denoting the three inner angles of $\triangle ACD$ by $x$, $y$, $z$, respectively, we have

Fig. 2
$$\angle AP_0C = 180^\circ - y = z + x.$$
Furthermore, from $B_1C_1 \perp P_0A$ and $B_1A_1 \perp P_0C$, we have $\angle B_1 = y$. In a similar way, we have $\angle A_1 = x$ and $\angle C_1 = z$. It follows that $\triangle A_1B_1C_1 \sim \triangle ABC$. Now let
$$B_1C_1 = \lambda BC, \quad C_1A_1 = \lambda CA, \quad A_1B_1 = \lambda AB.$$
Then for any point $M$ on the plane, we have
$$\begin{aligned} \lambda f(P_0) &= \lambda (P_0A \times BC + P_0D \times CA + P_0C \times AB) \\ &= P_0A \times B_1C_1 + P_0D \times C_1A_1 + P_0C \times A_1B_1 \\ &= 2S_{\triangle A_1B_1C_1} \\ &\le MA \times B_1C_1 + MD \times C_1A_1 + MC \times A_1B_1 \\ &= \lambda (MA \times BC + MD \times CA + MC \times AB) \\ &= \lambda f(M). \end{aligned}$$
That is to say, $f(P_0) \le f(M)$. Since $M$ is an arbitrary point, we know that $f(P)$ reaches the minimum at $P_0$, and at the same time $P_0A$, $P_0C$, $P_0D$ are concyclic. This completes the proof.

(2) From (1) we know that the minimum of $f(P)$ is
$$f(P_0) = \frac{2}{\lambda} S_{\triangle A_1 R_1 C_1} = 2\lambda S_{\triangle ABC}.$$
In the same way as is shown in the proof of (2) in Solution I, we find that both $\triangle ADC$ and $\triangle ABC$ are right-angled isosceles triangles; so we have $CD = AD = \frac{AC}{\sqrt{2}} = 1$, $AB = \sqrt{2}AC = 2$, $S_{\triangle ABC} = 1$,
$$BD = \sqrt{AB^2 + AD^2} = \sqrt{5}.$$
Furthermore, since $\triangle A_1B_1C_1 \sim \triangle ABC$, $\angle AB_1B = \angle AB_1C + \angle BB_1C = 90^\circ$, $AB_1BD$ is a rectangle. It follows that $B_1C_1 = BD = \sqrt{5}$. Therefore $\lambda = \frac{\sqrt{5}}{\sqrt{2}}$ and
$$$$
f(P)_{\min} = 2 \times \frac{\sqrt{5}}{\sqrt{2}} \times 1 = \sqrt{10}.

(1) We discuss the problem on the complex plane, and regard points $A$, $B$, $C$ as complex numbers. Then by the triangle inequality we have
$$| \vec{PA} \cdot \vec{BC} | + | \vec{PC} \cdot \vec{AB} | \ge | \vec{PA} \cdot \vec{BC} + \vec{PC} \cdot \vec{AB} |.$$
That is to say,
$$\begin{aligned} & |(A-P)(C-B)| + |(C-P)(B-A)| \\ \ge & |(A-P)(C-B) + (C-P)(B-A)| \\ = & |-P \times C - A \times B + C \times B + P \times A| \\ = & |(B-P)(C-A)| \\ = & |\vec{PB} \cdot \vec{AC}|. \end{aligned} \tag{1}$$
Then
$$\begin{aligned} & |\vec{PA} \cdot \vec{BC}| + |\vec{PC} \cdot \vec{AB}| + |\vec{PD} \cdot \vec{AC}| \\ \ge & |\vec{PB} \cdot \vec{AC}| + |\vec{PD} \cdot \vec{AC}| \\ = & (\vec{PB} + \vec{PD}) \cdot \vec{AC} \\ = & |\vec{BD}| \cdot \vec{AC}|. \end{aligned} \tag{2}$$
The equality in (1) holds only when the complex numbers $(A-P)(C-B)$ and $(C-P)(B-A)$ are in the same direction. This means that there is a real $\lambda > 0$ such that
$$(A-P)(C-B) = \lambda(C-P)(B-A).$$
That is to say,
$$\frac{A-P}{C-P} = \lambda \frac{B-A}{C-B}.$$
Therefore,
$$\arg\left(\frac{A-P}{C-P}\right) = \arg\left(\frac{B-A}{C-B}\right),$$
which means that the angle of rotation from $\vec{PC}$ to $\vec{PA}$ is equal to that from $\vec{BC}$ to $\vec{AB}$. It follows that $P$, $A$, $B$, $C$ are concyclic.
The equality in (2) holds only when $B$, $P$, $D$ are collinear and $P$ lies on the segment $BD$. This means that $f(P)$ reaches the minimum when $P$ lies on the circumscribed circle of $\triangle ABC$ and $P$, $A$, $B$, $C$ are concyclic.

(2) From (1) we know that $f(P)_{min} = BD \times CA$. The following steps are the same as those in the proof of (2) in Solution I.