= Solution
For <Banach spaces> $E,F$ and $1\le p<\infty$, define the weak $p$-norm of a finite family by
$$
w_p(x_1,\ldots,x_m)=\sup_{\phi\in B_{E^*}}\left(\sum_{i=1}^m|\phi(x_i)|^p\right)^{1/p}.
$$
A <bounded linear operator> $T:E\to F$ is an <absolutely p-summing operator> if $(\sum_i\|Tx_i\|^p)^{1/p}\le Cw_p(x_1,\ldots,x_m)$ for every finite family. The least constant is $\pi_p(T)$.
The <Pietsch factorization theorem> states that such an operator admits a <probability measure> $\mu$ on $K=B_{E^*}$ with its <weak-star topology> such that
$$
\|Tx\|\le\pi_p(T)\left(\int_K|\phi(x)|^p\,d\mu(\phi)\right)^{1/p}\qquad(x\in E).
$$
Equivalently, let $i:E\to C(K)$ be the evaluation embedding $i(x)(\phi)=\phi(x)$ and $j_p:C(K)\to L^p(\mu)$ the natural map. There is a <bounded linear operator> $V:H\to F$, where $H=\overline{j_pi(E)}\subset L^p(\mu)$, such that $T=Vj_pi$ and $\|V\|\le\pi_p(T)$. Conversely, any such factorization makes $T$ an <absolutely p-summing operator>, with $\pi_p(T)\le\|V\|$. The final operator is required on this closed subspace $H$; no extension to all of $L^p(\mu)$ is assumed.
Here is the existence proof. If $T=0$ the conclusion is immediate. Otherwise write $C=\pi_p(T)>0$. The <Banach-Alaoglu theorem> makes $K$ compact, and $\phi\mapsto|\phi(x)|^p$ is continuous for each $x\in E$. For finitely many vectors $x_1,\ldots,x_m$, put
$$
v(\phi)=(|\phi(x_i)|^p)_{i=1}^m,\qquad a=(C^{-p}\|Tx_i\|^p)_{i=1}^m.
$$
We claim the compact <convex hull> of $v(K)$ meets $a+[0,\infty)^m$. If it did not, finite-dimensional <Hahn-Banach separation theorem> would give weights $\lambda_i\ge0$ with
$$
\sup_{\phi\in K}\sum_i\lambda_i|\phi(x_i)|^p< C^{-p}\sum_i\lambda_i\|Tx_i\|^p.
$$
The weights are nonnegative because the second convex set is unbounded in every positive coordinate direction. This strict inequality contradicts the defining summing inequality applied to $\lambda_i^{1/p}x_i$. Thus a finite convex combination of the vectors $v(\phi)$ lies coordinatewise above $a$. Equivalently, a finitely supported <probability measure> satisfies all these finitely many domination constraints.
By the <Riesz-Markov-Kakutani representation theorem>, probability measures on compact $K$ are the positive norm-one functionals on $C(K)$ taking the constant function to one. They form a compact set in the <weak-star topology>, by the <Banach-Alaoglu theorem>. Each domination constraint $\int|\phi(x)|^p\,d\mu\ge C^{-p}\|Tx\|^p$ is closed. The finite feasibility just proved gives the <finite intersection property>, hence one measure $\mu$ satisfies every constraint.
Define $V(j_pi(x))=Tx$. The domination inequality makes this well-defined and bounded by $C$, and completeness of $F$ extends it to $H$. Conversely, any factorization satisfies
$$
\sum_i\|Tx_i\|^p\le\|V\|^p\int_K\sum_i|\phi(x_i)|^p\,d\mu(\phi)
\le\|V\|^p w_p(x_1,\ldots,x_m)^p.
$$
This proves both directions of the <Pietsch factorization theorem>, including the optimal factorization norm.
For the multiplication operator, let $f\in L^p(0,1)$ and take $g_1,\ldots,g_m\in C([0,1])$. Point evaluations are norm-one <linear functionals> on $C([0,1])$, so
$$
\sum_i\|fg_i\|_p^p
=\int_0^1|f(t)|^p\sum_i|g_i(t)|^p\,dt
\le\|f\|_p^p\sup_{t\in[0,1]}\sum_i|g_i(t)|^p
\le\|f\|_p^p w_p(g_1,\ldots,g_m)^p.
$$
Thus $M_f$ is an <absolutely p-summing operator> and $\pi_p(M_f)\le\|f\|_p$. Taking the single function $g=1$ gives the reverse inequality, so
$$
\boxed{\pi_p(M_f)=\|M_f\|=\|f\|_p.}
$$
For $f\ne0$, the explicit domination measure on $[0,1]$ is $|f(t)|^p\,dt/\|f\|_p^p$, pushed into $K$ by $t\mapsto\delta_t$. For $f=0$ everything is zero; no essential boundedness of $f$ is needed.
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