Answer
We have proven that $P$ lies on a sphere.
Work Step by Step
\[
P\left(z_{0}, y_{0}, x_{0}\right)=( a \cos \phi, a \sin \phi \sin \theta, a \sin \phi \cos \theta)
\]
Notice that
\[
\begin{aligned}
z_{0}^{2}+y_{0}^{2}+x_{0}^{2} &=a^{2} \cos ^{2} \phi+a^{2} \sin ^{2} \phi \sin ^{2} \theta+a^{2} \sin ^{2} \phi \cos ^{2} \theta \\
&=a^{2} \sin ^{2} \phi\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+a^{2} \cos ^{2} \phi \\
&=a^{2} \cos ^{2} \phi+a^{2} \sin ^{2} \phi \\
&=a^{2}\left(\cos ^{2} \phi+\sin ^{2} \phi\right) \\
&=a^{2}
\end{aligned}
\]
so we have used that $1=\sin ^{2} x+\cos ^{2} x$ for all real $x$.
Since $a^{2}=z_{0}^{2}+y_{0}^{2}+x_{0}^{2}, P$ lies on the sphere
\[
a^{2}=z^{2}+y^{2}+x^{2}
\]
Originally centered with radius $a=r$
