The answer seems to be "no" even for metrizable separable Banach spaces, which have property $\mathbf P$.

Take any infinite-dimensional separable Banach space $X$, fix a non-zero point $x_0\in X$ and a discontinuous linear functional $f:X\to\mathbb R$ such that $f(x_0)=1$. It is well-known that $f$ is not measurable (in many possible senses).

Define the bilinear function $m:X\times X\to X$ by $m(x,y)=f(x)\cdot f(y)\cdot x_0$. It can be show that the function $m$ is associative and $x_0$ is the unit of the multiplication $m$. The non-measurability of $f$ implies the non-measurability of the bilinear form $m$.

The answer seems to be "no" even for metrizable separable Banach spaces, which have property $\mathbf P$.

Take any infinite-dimensional separable Banach space $X$, fix a non-zero point $x_0\in X$ and a discontinuous linear functional $f:X\to\mathbb R$ such that $f(x_0)=1$. It is well-known that $f$ is not measurable (in many possible senses).

Define the bilinear function $m:X\times X\to X$ by $m(x,y)=f(x)\cdot f(y)\cdot x_0$. It can be show that the function $m$ is associative. The non-measurability of $f$ implies the non-measurability of the bilinear form $m$.