It would be misleading to write
$$\vec P=(r, \theta, \phi)$$ or
$$\vec P=\begin{pmatrix}r \\ \theta \\ \phi \end{pmatrix}$$
because the curvilinear coordinates $r,\theta,\phi$
don't behave like vector components with the usual simple rules for
vector addition and multiplication.
And it would easily lead to nonsensical conclusions
(as you already noticed).
It is better to imagine $\vec{P}$ with cartesian
components $x,y,z$ because these truly behave like
vector components. Using the definition of spherical
coordinates you have:
$$\vec{P}
=\begin{pmatrix}x \\ y \\ z\end{pmatrix}
=\begin{pmatrix}r\sin\theta\cos\phi \\
r\sin\theta\sin\phi \\
r\cos\theta \end{pmatrix} \tag{1}$$
$$\vec{P}'
=\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix}
=\begin{pmatrix}x+dx \\ y+dy \\ z+dz \end{pmatrix}$$
With this notation you get the basis vectors
$\vec e_\theta,\vec e_\phi,\vec e_r$
in a straight-forward way by differentiating (1).
$$\vec{e}_r = \frac{\partial\vec P}{\partial r}
=\begin{pmatrix}\sin\theta\cos\phi \\
\sin\theta\sin\phi \\ \cos\theta \end{pmatrix}$$
$$\vec{e}_\theta = \frac{\partial\vec P}{\partial \theta}
=\begin{pmatrix}r\cos\theta\cos\phi \\
r\cos\theta\sin\phi \\ -r\sin\theta \end{pmatrix}$$
$$\vec{e}_\phi = \frac{\partial\vec P}{\partial \phi}
=\begin{pmatrix}-r\sin\theta\sin\phi \\
r\sin\theta\cos\phi \\ 0 \end{pmatrix}$$
It is meant with the vectors above that you have equation
$$ \vec{P}'=\vec{P} + d\theta\ \vec{e}_\theta+d\phi\ \vec{e}_\phi+dr\ \vec{e}_r.$$
