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Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite* and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

*: Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras, and corresponds to the fact (proved in Murray and Von neuman original paper) that the sum of two orthogonal finite projection is again a finite projection.

Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite* and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

*: Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras, and corresponds to the fact (proved in Murray and Von neuman original paper) that in von Neuman algebras the sum of two orthogonal finite projection is again a finite projection.

Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite* and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

*: Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras, and corresponds to the fact (proved in Murray and Von neuman original paper) that the sum of two orthogonal finite projection is again a finite projection.

Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite and any projection of a finite factor is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.

Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent.

This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator.

One now has the following lemma:

Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$

This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite.

Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand.