Ethyl lithium is a strong base, which means it will deprotonate the carboxylic acid.
PhCO2H + EtLI --> PhCO2Li + Et-H
This lithium salt can be attacked by another equivalent of ethyl lithium, which forms PhC(OLi)2-Et. When you learned about Grignards, you probably learned that Grignards add twice to esters to form tertiary alcohol, and that the second addition is impossible to stop. This reaction is a way of overcoming that. Unlike the PhC(OR)(OMg)Et tetrahedral intermediate, which collapses to eliminate ROMg and produces PhCOEt, the PhC(OLi)2Et intermediate does not collapse to form a ketone because OLi is not a leaving group.
Addition of acid protonates both oxygens. This intermediate - PhC(OH)2Et - is called a hydrate, and under acidic conditions, will form a ketone (in this case phenyl ethyl ketone/1-phenylpropanone/propiophenone).
I'm not sure what the PCC will do at this point. PCC oxidizes alcohols to ketones (and under certain conditions aldehydes to carboxylic acids). In fact I would go so far as to say PCC does nothing here.