Given $P\left(2 t^{2}, 2 t\right), M(1,0)$, the line through $P$ is $y-2 t=k\left(x-2 t^{2}\right)$ and it is tangent to the circle $M$, then we have
$$\begin{array}{l} \frac{\left|k\left(1-2 t^{2}\right)+2 t\right|}{\sqrt{1+k^{2}}}=1 \\ \Rightarrow 4 t^{2}\left(t^{2}-1\right) k^{2}-4 t\left(2 t^{2}-1\right) k+4 t^{2}-1=0 \end{array}$$

Let the slopes of the lines $P B, P C$ be $k_{1}, k_{2}$, then
$$\begin{array}{l} y_{B}=2 t-2 t^{2} k_{1}, y_{C}=2 t-2 t^{2} k_{2}, \\ S_{\triangle P B C}=\frac{1}{2} \cdot 2 t^{2}\left|y_{B}-y_{C}\right|=2 t^{4}\left|k_{1}-k_{2}\right| \\ =2 t^{4} \cdot \frac{\sqrt{16 t^{2}\left(2 t^{2}-1\right)^{2}-16 t^{2}\left(t^{2}-1\right)\left(4 t^{2}-1\right)}}{4 t^{2}\left(t^{2}-1\right)} \\ =2 t^{2} \cdot \frac{|t| \sqrt{\left(2 t^{2}-1\right)^{2}-\left(t^{2}-1\right)\left(4 t^{2}-1\right)}}{t^{2}-1} \\ =\frac{2 t^{4}}{t^{2}-1}=2\left(\frac{t^{4}-1+1}{t^{2}-1}\right)=2\left(t^{2}+1+\frac{1}{t^{2}-1}\right) \\ =2\left(t^{2}-1+\frac{1}{t^{2}-1}+2\right) \geqslant 2(2+2)=8, \end{array}$$

Equality holds when $t^{2}=2 \Rightarrow t= \pm \sqrt{2}$. Therefore, the minimum area of $\triangle P B C$ is 8.

15 Let $P\left(x_{0}, y_{0}\right) 、 B(0, b) 、 C(0, c)$, and assume $b>c$. The equation of the line $P B$ is
$$y-b=\frac{y_{0}-b}{x_{0}} x$$

Simplifying, we get
$$\left(y_{0}-b\right) x-x_{0} y+x_{0} b=0.$$

Since the distance from the center $(1,0)$ to $P B$ is 1, we have
$$\frac{\left|y_{0}-b+x_{0} b\right|}{\sqrt{\left(y_{0}-b\right)^{2}+x_{0}^{2}}}=1$$

Thus,
$$\left(y_{0}-b\right)^{2}+x_{0}^{2}=\left(y_{0}-b\right)^{2}+2 x_{0} b\left(y_{0}-b\right)+x_{0}^{2} b^{2},$$

Given that $x_{0}>2$, the above equation simplifies to
$$\left(x_{0}-2\right) b^{2}+2 y_{0} b-x_{0}=0,$$

Similarly,
$$\left(x_{0}-2\right) c^{2}+2 y_{0} c-x_{0}=0.$$

Therefore,
$$b+c=\frac{-2 y_{0}}{x_{0}-2}, \quad b c=\frac{-x_{0}}{x_{0}-2},$$

Then,
$$(b-c)^{2}=\frac{4 x_{0}^{2}+4 y_{0}^{2}-8 x_{0}}{\left(x_{0}-2\right)^{2}}$$

Since $P\left(x_{0}, y_{0}\right)$ is a point on the parabola, we have $y_{0}^{2}=2 x_{0}$, thus
$$(b-c)^{2}=\frac{4 x_{0}^{2}}{\left(x_{0}-2\right)^{2}},$$

which gives
$$b-c=\frac{2 x_{0}}{x_{0}-2}$$

Therefore,
$$\begin{aligned} S_{\triangle P B C} & =\frac{1}{2}(b-c) \cdot x_{0} \\ & =\frac{x_{0}}{x_{0}-2} \cdot x_{0} \\ & =\left(x_{0}-2\right)+\frac{4}{x_{0}-2}+4 \\ & \geqslant 2 \sqrt{4}+4 \\ & =8, \end{aligned}$$

Equality holds when $\left(x_{0}-2\right)^{2}=4$, i.e., $x_{0}=4, y_{0}= \pm 2 \sqrt{2}$. Therefore, the minimum value of $S_{\triangle P B C}$ is 8.