Does secure multiplication makes sense in the case of 2 inputs? I am looking at a protocol to securely multiply numbers, so as to preserve privacy of inputs. 

In particular, suppose there are three parties $P_1$, $P_2$ and $P_3$ and 2 numbers to multiply, $a \in \mathbb{F}$ and $b \in \mathbb{F}$ such that $\mathbb{F}$ is a finite field of prime order $p$. More precisely, $a$ is the $P_1$'s input, $b$ is $P_2$'s input and $P_3$ has no input. $P_1$ and $P_2$ shares their input among the parties using 2-out-of-3 secret sharing and, in this way, each party can compute a share of the product. Then, shares of the product are securely summed and all the parties can compute the result.

In this aim, I am wondering whether does it make sense when we have only two parties?

If $p$ is big enough (such that $a \cdot b &lt; p$) , once $P_1$ knows $a \cdot b \mod p$, by knowing $a$, it could easily compute the secret value $b$.

For this reason, I do not see the point of using secure multiplication in this case. Could somebody help me understand this?
There is no way to avoid that a party infers knowledge about another party input by looking at the result and his or her input. This is not considered a security issue of the protocol.