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(C) $1$. The starting material is $Ph-CH(OH)-C \equiv C-CH=CH-OMe$. $2$. Treatment with $H_2 / Pd-BaSO_4$ (Lindlar's catalyst) selectively reduces the

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$Ph-(CH=CH)_2-CHO$

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(C) $1$. The starting material is $Ph-CH(OH)-C \equiv C-CH=CH-OMe$. $2$. Treatment with $H_2 / Pd-BaSO_4$ (Lindlar's catalyst) selectively reduces the alkyne to a cis-alkene,forming intermediate $(A)$: $Ph-CH(OH)-CH=CH-CH=CH-OMe$. $3$. Subsequent treatment with $H_3O^+$ leads to the hydrolysis of the enol ether. The protonation of the double bond followed by resonance and water attack leads to the formation of an aldehyde. $4$. The final product $(B)$ is $Ph-CH=CH-CH=CH-CHO$,which can be written as $Ph-(CH=CH)_2-CHO$.

Product $(B)$ is

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