Let $f : U\subset \mathbb{R}^3\to \mathbb{R}$ be given by $f(x,y,z)=\cos x\sin y e^z$ then we have that your surface is indeed a level set $M = f^{-1}(0)$. Then it's easy: remember that the gradient of a function is orthogonal to the level sets. Using this we have that
$$\nabla f(x,y,z)=(-\sin x\sin ye^z, \cos x\cos y e^z, \cos x\sin y e^{z})$$
So that at $(\pi/2,1,0)$ we have $\nabla f(\pi/2,1,0)=(-\sin 1,0,0)$, so that the normal is a multiple of the vector $e_1$, and hence since the magnitude of the normal vector doesn't matter, we can pick the normal vector to be $e_1$. Of course then, the tangent plane is just the $yz$ plane.
After applying a rigid motion to $S$ and $P$, we may as well assume $(x_0,y_0,z_0)=(0,0,0)$ and $P$ is the $xy$-plane. Let $U$ be a small neighborhood of the origin in $S$; by taking $U$ to be small enough, we may assume that $U$ is the image of an open disk in $\mathbb R^2$ under a local parametrization, and thus $U$ is homeomorphic to a disk. The hypothesis implies that the origin is the only point of $U$ for which the $z$-coordinate is zero, so $U\smallsetminus\{(0,0,0)\}$ is contained in the set where $z\ne 0 $. Since a disk minus a point is connected, it follows that either $z>0$ on all of $U\smallsetminus\{(0,0,0)\}$ or $z<0$ on all of $U\smallsetminus\{(0,0,0)\}$. After reflecting in the $xy$-plane, we may assume it is $z>0$.
Now suppose $v$ is a tangent vector to $S$ at the origin. Then there is a smooth curve $c:(-\varepsilon,\varepsilon)\to U$ satisfying $c(0) = (0,0,0)$ and $c'(0) = v$. If we write $c(t) = (x(t),y(t),z(t))$, we see that $z(t)$ attains a minimum at $t=0$, and therefore $z'(0)=0$. This means that every tangent vector at the origin has zero $z$-coordinate, so $T_{(0,0,0)}S$ is contained in the $xy$-plane (i.e., in $P$). Since both $T_{(0,0,0)}S$ and $P$ are two-dimensional, they must be equal.
Best Answer
If a surface is defined by a regular function $z=g(x,y)$, the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ in terms of the partial derivatives of $g$ is
$$z=z_{0}+\left. \frac{\partial g}{\partial x}\right\vert _{(x_{0},y_{0})}(x-x_{0})+\left. \frac{\partial g}{\partial y}\right\vert _{(x_{0},y_{0})}(y-y_{0}).$$
If the surface is defined implicitly by $f(x,y,z)=0$, then $z=g(x,y)=f^{-1}(0)$ (i.e. $f(x,y,g(x,y))\equiv 0$). Since
$$\frac{\partial g}{\partial x}=-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}$$
and
$$\frac{\partial g}{\partial y}=-\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z},$$
the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ is given by
$$z=z_{0}-\left(\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(x-x_{0})-\left(\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(y-y_{0})$$
or
$$(x-x_{0})\left(\frac{\partial f}{\partial x}\right)_{(x_{0},y_{0},z_{0})}+(y-y_{0})\left(\frac{\partial f}{ \partial y}\right)_{(x_{0},y_{0},z_{0})}+(z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}=0.$$
In compact notation, $\mathbf{x}=\left( x,y,z\right) ,\mathbf{x}_{0}=\left( x_0,y_0,z_0\right) $, we get
$$\left( \mathbf{x}-\mathbf{x}_{0}\right) \cdot \mathbf{\nabla }f(\mathbf{x}_{0})=0.$$