Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)?
In,Thermodynamics,Work is defined as 
$W$= $-\int_{V_1}^{V_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V_1=V_2$.

Coming to the enthalpy,you are right.
$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,
$\Delta$H=$\Delta$U + $P$$\Delta$V + $V$$\Delta$P.
At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\Delta$V.But,according,to first law of Thermodynamics,$Q$= $\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So,$W$= $-p\int_{V_1}^{V_2}dV$= -$p$($V_2-V_1$) = -$p$$\Delta$V.So, first law in constant presuure becomes,$Q_p$= $\Delta$U - (-$p$$\Delta$V)= $\Delta$U+$p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.