Solution

Reflection from a perfect metal is modeled using a fictitious wave that, inside the metal, propagates toward the surface. Outside the metal, there exists the real incident wave; inside, the fictitious wave moves in the opposite direction. By superposing both, the tangential electric field vanishes at the surface (plane $AB$). This fictitious wave, upon crossing the boundary, becomes the real reflected wave.

Choice of geometry and waves:
Metal surface: plane $x = 0$. The metal occupies $x > 0$; vacuum occupies $x < 0$.
Incident wave (traveling toward $+x$):
$$\vec{E}_i = E_0 \cos(\omega t - kx)\hat{y}, \quad \vec{B}_i = \frac{E_0}{c} \cos(\omega t - kx)\hat{z}$$
Reflected wave (real, traveling toward $-x$):
$$\vec{E}_r = -E_0 \cos(\omega t + kx)\hat{y}, \quad \vec{B}_r = \frac{E_0}{c} \cos(\omega t + kx)\hat{z}$$
The sign reversal of $\vec{E}_r$ ensures that the total tangential component vanishes at $x = 0$.

Electromagnetic field outside the metal ($x < 0$):
Superposing the incident and reflected waves creates a standing wave:
$$\vec{E}(x, t) = E_0 [\cos(\omega t - kx) - \cos(\omega t + kx)]\hat{y} = 2E_0 \sin(kx) \sin(\omega t)\hat{y}$$
$$\vec{B}(x, t) = \frac{E_0}{c} [\cos(\omega t - kx) + \cos(\omega t + kx)]\hat{z} = \frac{2E_0}{c} \cos(kx) \cos(\omega t)\hat{z}$$

Instant when the crest of the incident wave reaches the surface:
The crest (positive maximum of $\vec{E}_i$) arrives at $x = 0$ when $\cos(\omega t) = 1$, i.e., $\omega t = 0, 2\pi, \dots$
Taking $t = 0$:
$$\vec{E}(x, 0) = 0 \quad \text{(zero everywhere outside the metal)}$$
$$\vec{B}(x, 0) = \frac{2E_0}{c} \cos(kx)\hat{z}$$
Near the metal plane itself ($x \to 0^-$), the magnetic field induction is maximal and equals $\vec{B} = \frac{2E_0}{c}\hat{z}$.