Yes, these topologies agree basically by definition once you understand the isomorphism $M(A)\cong \mathcal L(A)$ (since $tθ_{a,b}=θ_{t(a),b}$ the canonical isomorphism $A≅ \mathcal K(A)$ takes this element to $t(a)b^*=t(ab^*)$). The isomorphism $A \to \mathcal K(A)$ is given by $a \mapsto \theta_{a_1,a_2}$ where $a=a_1a_2^\ast$ is any way of writing $a$ as a product of two elements (any element in a $C^\ast$-algebra can be written as a product of two elements, e.g. if a $a= u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast\ast}$, then $a_1 := u|a|^{1/2} \in A$ and $a = a_1 |a|^{1/2}$).

Yes, these topologies agree basically by definition once you understand the isomorphism $M(A)\cong \mathcal L(A)$ (since $tθ_{a,b}=θ_{t(a),b}$ the canonical isomorphism $A≅ \mathcal K(A)$ takes this element to $t(a)b^*=t(ab^*)$). The isomorphism $A \to \mathcal K(A)$ is given by $a \mapsto \theta_{a_1,a_2}$ where $a=a_1a_2^\ast$ is any way of writing $a$ as a product of two elements (any element in a $C^\ast$-algebra can be written as a product of two elements, e.g. if a $a= u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast\ast}$, then $a_1 := u|a|^{1/2} \in A$ (this can be seen by approximating $t\mapsto t^{1/2}$ by polynomials) and $a = a_1 |a|^{1/2}$).