vapor pressure of one part of binary compound
No. The equilibrium vapor (gas) pressure is defined by the corresponding equilibrium. When it's just the element $\ce{B}$ that evaporates, the equilibrium is $$\ce{B(liquid)&lt;=&gt;B(gas),}$$ and the thermodynamics of evaporation (or sublimation if $\ce{B}$ is solid), such as the Clausius–Clapeyron equation, gives you the equilibrium $p-T$ relationship.

When you have nonstoichiometric compound $\ce{AB_{1-x}}$ which loses $\ce{B}$, the process is quite different. Let's imagine purely hypothetical $\ce{AB}$, in the crystal structure of which neutrally charged $\ce{B}$ vacancies, $\ce{V^x_B}$, can be formed. Then, our hypothetical equilibrium between the gas phase and defects in solid can be written in Kröger–Vink notation as $$\ce{B^x_{B}&lt;=&gt;V^x_{B} + B_{gas}}.$$ In this case, until $\ce{AB_{1-x}}$ decomposes too much, the equilibrium constant $K=\frac{p_B\cdot[\ce{V^x_{B}}]}{[\ce{B^x_{B}}]}$ and its temperature dependence define the "partial pressure of $\ce{B}$ - temperature - composition" relationship for $\ce{AB_{1-x}}$.

Please note that the example above is oversimplified for clarity. In the real-world applications of defect chemistry, point defects in nonstoichiometric solids are often charged, and they tend to interact with each other, complicating the equilibria.