= Solution
The set $K\cap P$ is compact because $K$ is closed and bounded, and the product
$$
G(x)=\prod_{j=1}^n x_j
$$
is continuous. By the <extreme value theorem>, $G$ has a maximizer. Since $K$ meets the interior of the <nonnegative orthant> $P$, some feasible product is positive, so every maximizer lies in $\operatorname{Int}P$.
On $\operatorname{Int}P$, maximizing $G$ is equivalent to maximizing
$$
F(x)=\log G(x)=\sum_{j=1}^n\log x_j.
$$
The Hessian of $F$ is
$$
D^2F(x)=\operatorname{diag}\left(-\frac1{x_1^2},\ldots,-\frac1{x_n^2}\right),
$$
which is negative definite. Thus $F$ is a <strictly concave function>, and it has at most one maximizer on the <convex set> $K\cap P$. Denote this unique point by $x^*$.
For any $x\in K\cap P$, the line segment
$$
x(t)=(1-t)x^*+tx,\qquad 0\leq t<1,
$$
lies in $K\cap\operatorname{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:
$$
0\geq\left.\frac d{dt}F(x(t))\right|_{t=0}
=\sum_{j=1}^n\frac{x_j-x_j^*}{x_j^*}.
$$
Therefore
$$
\boxed{\sum_{j=1}^n\frac{x_j}{x_j^*}\leq n}.
$$
This is the supporting inequality for the <product maximizer on a compact convex subset of the positive orthant>.
Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^*$. Hence
$$
x^*=(r,\ldots,r)
$$
for some $r>0$. More explicitly,
$$
\boxed{
r=\max_{x\in K\cap P}\frac1n\sum_{j=1}^n x_j
}.
$$
Indeed, <cyclic symmetry averaging> puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^*=r$ gives $\sum_jx_j\leq nr$.
Finally, given $a=(a_1,\ldots,a_n)\in\operatorname{Int}P$, define
$$
K=\left\{x\in P:
\sum_{j=1}^n\frac{x_j}{a_j}\leq n
\right\}.
$$
This set is nonempty, closed, convex and bounded, and it contains $a$. By the <arithmetic-geometric mean inequality>,
$$
\frac{\prod_jx_j}{\prod_ja_j}
=\prod_{j=1}^n\frac{x_j}{a_j}
\leq
\left(\frac1n\sum_{j=1}^n\frac{x_j}{a_j}\right)^n
\leq1.
$$
Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has
$$
\boxed{x^*=a}.
$$
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