Let $M$ be a von Neumann algebra and $\varphi$ is a faithful normal state on $M$. Suppose that $M_{\varphi}$ is a type II$_1$ factor.
Suppose that if $p, q$ are two nonzero projections in $M_{\varphi}$, we can find $v\in M$ such that $vv^*=p, v^*v=q$ and $v\varphi=\frac{\varphi(p)}{\varphi(q)}\varphi v$.
If $\lambda\geq 1$, choose $n\geq 1$ such that $\frac{1}{n}\leq\frac{1}{\lambda}$. Then we can find a partition of unity $q_1,\cdots,q_n$ in $M_{\varphi}$ and some projections $p_1,\cdots, p_n$(not necessarily orthogonal) such that $\lambda^{-1}\varphi(p_k)=\varphi(q_k)$.
How to construct $p_i$ and $q_i(i=1,\cdots,n)$? If $q_1+q_2+\cdots q_n=1$, why does there exist some projections $p_1,\cdots, p_n$(not necessarily orthogonal) such that $\lambda^{-1}\varphi(p_k)=\varphi(q_k)$?
I tried to construct $q_i$ in the following way: choose $0\leq a\in M_{\varphi}$ and let $B_1\cup B_2\cdots \cup B_n=\sigma(a)$.Set $q_i=1_{B_i}(a)$, then $q_1+q_2+\cdots q_n=1$. But I don't know how to construct each $p_i$.
As stated in your third paragraph, the problem may have no solution. For example take $n=\lambda=2$, and $q$ a projection with $\varphi(q)=\frac23$. Then the equality $\varphi(p)=\lambda\varphi(q)$ gives $\varphi(p)=\frac43$, which is impossible.
If one gets to choose both the $q$ and the $p$, then you can do the following. Since $\varphi$ is faithful and normal and $M_\varphi$ is a II$_1$-factor, given any projection $q$ and $0\leq t\leq \varphi(q)$, there exists $p\leq q$ with $\varphi(p)=t$.
Fix $m\in\mathbb N$ with $m\geq\lambda$ and let $q_1,\ldots,q_m$ be pairwise orthogonal projections with $q_1+\cdots+q_m=1$ and $\varphi(q_j)=\frac1m$. For all $j$.
Fix $q_j$. We have $\varphi(1-q_j)=1-\frac1m$. Since $$\frac{\lambda-1}m\leq\frac{m-1}m=1-\frac1m$$ there exists a projection $r_j\leq 1-q_j$ with $\varphi(r_j)=\frac{\lambda-1}m$. Then $p_j=q_j+r_j$ is a projection and $$ \varphi(p_j)=\varphi(q_j)+\varphi(r_j)=\frac1m+\frac{\lambda-1}m=\frac\lambda m=\lambda\,\varphi(q_j). $$