Why does no Robinson Annulation occur after the Michael addition?
As illustrated in the following figure, the Robinson annulation reaction involves two steps.  The first step involves the Michael addition of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below).  In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an intramolecular aldol condensation to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.



Often in Robinson annulations methyl vinyl ketone is used. It is important that  the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl so that the enolate required in the second step can be formed. 


  Explain why no Robinson annulation cyclic product is observed?


In your example, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your Product C can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.